283 111 23MB
English Pages [540] Year 2019
Physics Galaxy Volume IV
Optics & Modern Physics Ashish Arora Mentor & Founder
PHYSICSGALAXY.COM
World's largest en(yclopedia ofonline video lectures on High School Physics
G K Publications (P) Ltd
CL MEDIA (P) LTD. First Edition 2000 Revised Edition 2017
© AUTHOR
Administrative and Production Offices
No part ofthis book may be reproduced in a retrieval system or transmitted, in any form or by any means, electronics, mechanical, photocopying, recording, scanning and or without the written permission of the author/publisher.
Published by : CL Media (P) Ltd. KH No. 1027, Ramnagar, Salempur, Rajputana, Roorkee (U.K.)247667^ Marketedby : G.K. Publications (P) Ltd.
ISBN
:
A41, Lower Ground Floor, Espire Building, Mohan Cooperative Industrial Area,
9788183558938
Main Mathura Road,
Typeset by : CL Media DTP Unit
New Delhi  110044
For product information :
Visit www.gkpuhlication8»com or email to [email protected]
My beloved wife
My Parents, Son, Daughter
Dedicated to
In his teaching career since 1992 Ashish Arora personally mentored more than 10000 IITians and students who reached global heights in various careerand profession chosen. It is his helping attitude toward studentswith which all his studentsrentember him in life for his contribution in their success and keep connections with him live. Below is the list of some of the successful students in International Olympiad personally taught by him. NAVNEET LOrWAL
International GQLD Medal in IPbO2000 at LONDON, Also secured AIR4 in IIT JEE 2000 PROUD FOR INDIA ; Navneet Loiwal was the first Indian Student who won first International GOLD Medal
for our country In International Physics Olympiad. DUNGRA RAM CHOUDHARY
HARSHIT CHOPRA KUNTAL LOYA LUV KUMAR
RAJHANS SAMDANI.
SHANTANU BHARDWAJ
SHALEEN HARLAlKA
AIR1 in IIT JEE 2002
National Gold Medal in INPbO2002 and got AIR2 in IIT JEE2002 A Girl Student got position AIR8 in IIT JEE 2002 National Gold Medal in INPbO2003 and got AIR3 in IIT JEE2003 National Gold Medal in INPbO2003 and got AIR5 in IIT JEE2003 International SILVER Medal in IPbO2002 at INDONESIA
International GOLD Medal in IPb0T2003 at CHINA and got AIR46 in IIT JEE2003
TARUN GUPTA
National GOLD Medal in INPbO2005
APEKSHA KHANDELWAL
National GOLD Medal in INPbO2005
ABHINAV SINHA
Honble Mension Award in APbO2006 at KAZAKHSTAN
RAMAN SHARMA
International GOLD Medal in IPbO2007 at IRAN and got AIR20 in IIT JEE2007
PRATYUSH PANDEY
International SILVER Medal in IPhO2007 at IRAN and got AIR85 in IIT JEE2007
GARVIT JUNIWAL
International GOLD Medal in IPbO2008 at VIETNAM and got AIR10 in IIT JEE2008
ANKIT PARASHAR
National GOLD Medal in INPbO2008
HEMANT NOVAL ABHISHEK MITRUKA SARTHAK KALANI
National GOLD Medal in INPbO2008 and got AIR2S in IIT JE&2008 National GOLD Medal in INPbO2009 National GOLD Medal in INPbO2009
ASTHA AGARWAL
International SILVER Medal in IJSO2009 at AZERBAIJAN
RAHUL GURNANI
International SILVER Medal in IJSO2009 at AZERBAIJAN
AYUSH SINGHAL
International SILVER Medal in IJSO2009 at AZERBAIJAN
MEHUL KUMAR
ABHIROOP BHATNAGAR AYUSH SHARMA
International SILVER Medal in IPhO2010 at CROATIA and got AIR19 in IIT JEE2010 National GOLD Medal in INPbO2010
International Double GOLD Medal in IJSO2010 at NIGERIA
AASTHA AGRAWAL
Honble Mension Award in APbO2011 at ISRAEL and got AIR93 in IIT JEE 2011
ABHISHEK BANSAL
National GOLD Medal in INPbO2011
^ SAMYAK DAGA
National GOLD Medal in INPbO2011
SHREY GOYAL
National GOLD Medal in INPbO2012 and secured AIR24 in IIT JEE 2012
RAHUL GURNANI
National GOLD Medal in INPbO2012
JASPREET SINGH JHEETA
National GOLD Medal in INPbO2012
DIVYANSHU MUND
SHESHANSH AGARWAL
National GOLD Medal in INPbO2012 International SILVER Medal in IAO2012 at KOREA
SWATI GUPTA
International SILVER Medal in IJSO2012 at IRAN
PRATVUSH RAJPUT
' International SILVER Medal in IJSO2012 at IRAN
SHESHANSH AGARWAL
International BRONZE Medal in IOAA2013 at GREECE
SHESHANSH AGARWAL
International GOLD Medal in IOAA2014 at ROMANIA
SHESHANSH AGARWAL
International SILVER Medal in IPbO2015 at INDIA and secured AIR58 in JEE(AdTanced)20I5
VIDUSHI VARSHNEY
International SILVER Medal in IJSO2015 to be held at SOUTH KOREA
AlvIAN BANSAL
AIR1 in JEE Advanced 2016
KUNAL GOYAL
AIR3 in JEE Advanced 2016
GOURAV DIDWANIA DIVYANSH GARG
AIR9 in JEE Advanced 2016 International SILVER Medal in IPbO2016 at SWITZERLAND
ABOUT THE AUTHOR
The complexities ofPhysics have given nightmares to many, but the homegrown genius ofJaipur, Ashish Arora has helped several students to live their dreams by decoding it. Newton Law ofGravitation and Faraday's Magnetic force ofattraction applyperfectlywell with this unassuming genius. APied Piper ofstudents, his webportal https;//www.physicsgalaxy.com. The world slargest encyclopedia ofvideo lectures on high school Physics possesses strong gravitational pull and magnetic attraction for students who want to make itbig inlife.
Ashish Arora, gifted with rare ability to train masterminds, has mentored over 10,000 IITians in his past 24 years ofteaching sojourn including lots ofstudents made it toTop 100 in nTJEE/JEE(Advance)includingAIRl andmanyinToplO. Apart from
that, he has also groomed hundreds ofstudents for cracking International Physics Olympiad. No wonder his student Navneet Loiwal brought laurel to the country bybecoming the first Indian to win aGold medal at the 2000 International Physics Glvmniad in London (UK).
His special ability to simplify the toughest ofthe Physics theorems and applications rates him as one among the best Physics teachers in the world. With this, Arora simply defies the logic that perfection comes with age. Even at 18 when he started teaching Physics while pursuing engineering, he was as engaging as he is now. Experience, besides graying his hair, has just widened his horizon.
Now after encountering all tribes ofstudents  some brilliant and some notsointelligent this celebrated teacher has embarked
upon anoble mission to make the entire galaxy ofPhysics inform ofhis webportal PHYSICSGALAXY.COM to serve and help global students in the subject. Today students from 221 countries are connected with this webportal. On any topic ofphysics students can post their queries in INTERACT tab ofthe webportal on which many global experts with Ashish Arora reply to several queries posted online by students.
Dedicated to global students ofmiddle and high school level, his website www.physicsgalaxy.com also has teaching sessions dubbed inAmerican accent and subtitles in 87 languages. For students in India preparing for JEE &NBET, his online courses will be available soon on PHYSICSGALAXY.COM.
FOREWORD
It has beenpleasurefor me to follow the progressEr.AshishArorahas made in teachingand professional career. In the last about two decades he has actively contributed in developing several new techniques for teaching & learning of Physics and driven important contribution to Science domain throughnurturingyoung students and budding scientists. Physics Galaxyis onesuch example ofnumerous efforts he has undertaken.
The 2nd edition of Physics Galaxyprovides a good coverage of various topics of Mechanics, Thermodynamics and Waves, Optics & Modem Physics and Electricity & Magnetism through dedicated volumes. It would be an important resource for students appearing in competitiveexaminationfor seeking admissionin engineering and medicalstreams. "Eversion" of the book is also being launched to allow easy access to all.
The structure ofbook islogical and thepresentation is innovative. Importantly the book covers someofthe concepts onthe basis of realistic experiments and examples. The book has been written in an informal style to help students leara faster and more interactively with better diagrams and visual appeal of the content. Each chapter has variety of theoretical and numerical problems to test the knowledge acquired by students. The book also includes solution to all practice exercises with several new illustrations and problems for deeper leaming. 1am sure the book will widen the horizons of knowledge in Physics and will be found very useful by the students for developing indepth understanding of the subject.
May 13, 2017 Prof. Sandeep Sancheti Ph. D. (U.K.). B.Tech. FIETE, MIEEE
President Manipal University Jaipur
PREFACE
For a science student, Physics is the most important subject, unlike to other subjects it requires logical reasoning and high imagination of brain. Without improvingthe level ofphysics it is very difficultto achievea goal in the present age of competitions. To score better, one does not require hard working at least in physics. It just requires a simple understanding and approach to think a physicalsituation.Actuallyphysicsis the surrounding of our everydaylife.All the six parts of generalphj^icsMechanics, Heat, Sound, Light, Electromagnetism and Modem Physics are the constituents of our surroundings. If you wish to make the concepts of physics strong, you should try to understand core concepts of physics in practical approach rather than theoretical. Whenever you try to solve a physics problem, first create a hypothetical approach rather than theoretical. Whenever you try to solve a physics problem, first create a hypothetical world in your imagination about the problem and try to think psychologically,
whatthe nextstepshould be, the bestanswer would begiven byyourbrainpsychology. Formaking physics strongin all respects and you should try to merge and understand all the concepts with the brain psychologically. The book PHYSICS GALAXY is designed in a totally different and fi"iendly approach to develop the physics concepts psychologically. The book is presented in four volumes, which covers almost all the core branches of general physics. First volume covers Mechanics. It is the most important part of physics. The things you will leam in this book will form a major foundation for understanding of other sections of physics as mechanics is used in all other branches of physics as a core fundamental. In this bookeverypart of mechanicsis explainedin a simpleand interactive experimental way. The bookis divided in seven major chapters, covering the complete kinematics and dynamics of bodieswith both translational and rotational motion. then gravitation and completefluid staticsand dynamics is coveredwith several applications.
The best way of understanding physics is the experimentsand this methodology I am using in my lectures and I found that it helps students a lot in concept visualization. In this book I have tried to translatethe things as I used in lectures. After every important section thereare several solved examples included withsimple andinteractive explanations. It mighthelpa student in a way that the student does not require to consult any thing with the teacher. Everything is self explanatory and in simple language.
Oneimportant factor in preparation ofph>^ics I wish to highlight thatmost ofthe student after reading thetheory ofa concept. start working out the numerical problems. This is not the efficient wayof developing concepts in brain. Toget the maximum benefit ofthebook students should readcarefully thewhole chapter at least three or four times with allthe illustrative examples andwithmorestress onsomeillustrative examples included in thechapter. Practice exercises included afterevery theorysection in each chapter is for the purpose of indepth understanding of the applications of concepts covered. Illustrative examples are explaining some theoretical concept in theform ofanexample. After a thorough reading ofthechapter students can startthinking on discussion questionsand start working on numerical problems. Exercises given at the end of each chapter are for circulation of all the concepts in mind. There are two sections, first is the discussion questions, which aretheoretical andhelp inxmderstanding the concepts atrootlevel. Second section is ofconceptual MCQs which helpsin enhancing the theoretical thinkingofstudents and building logical skillsin the chapter. Third section of numerical MCQs helpsinthe developing scientific and analytical application of concepts. Fourth section ofadvance MCQs with one or more optionscorrecttypequestions is for developing advanceand comprehensive thoughts. Last sectionis the Unsolved Numerical Problemswhich includessome simpleproblems and some tough problemswhich require the building fundamentals of physicsfrombasics to advancelevelproblemswhich are usefulin preparationof NSEP, INPhOor IPhO.
In this second edition of the book I have included the solutions to all practiceexercises, conceptual, numerical and advance MCQs to support students who are dependent on theirselfstudy and not getting access to teachers for their preparation.
This book has taken a shape just because of motivational inspiration bymy'mother 20 years agowhen I just thought to write something for my students. She always motivated and was on my side whenever I thought to develop some new learning methodology for my students.
rdon't have words for mybest friend mywife Anuja for always being together with me to complete this book in the unique style and format.
I would like to pay my gratitude to Sh. Dayashankar Prajapati in assisting me to complete the task in Design Labs of PHYSICSGALAXY.COM and presenting the book in totallynew format ofsecond edition.
At last but the most important person, my father who has devoted his valuable time to finally present the book in such aformat and a simple language, thanks isa very small word for hisdedication inthis book.
In this second edition Ihave tried my best to make this book error free but owing to the nature ofwork, inadvertently, there is possibilityoferrors leftuntouched. Ishall be grateful to the readers, iftheypoint out me regarding errors and oblige me bygiving their valuable and constructive suggestions via emails for further improvement ofthe book.
Date:May,2017
Ashish Arora PHYSICSGALAXY.COM
B80, Model Town, Malviya Nagar, Jaipur302017 emails: [email protected] [email protected]
CONTENTS Atomic Physics
_
^
•
A
1.1 A Brief History to Atomic Physics 1.2 Thomson's Atomic Model
1.3 Ruthorford's Atomic Model
^
1.4 Bohr's Model ofanAtom
^
1.4.1 First Postulate
^
1.4.2 Second Postulate
1.4.3 ThirdPostulate
•
"
^
1.5 Properties ofElectron in Bohr's Atomic Model 1.5.1 Radius of nth Orbit in Bohr Model 1.5.2 Velocity ofElectron in nth Bohr's Orbit 1.5.3 Angular Velocity ofElectron in nth Bohr's Orbit 1.5.4 Frequency ofElectron in nth Bohr's Orbit 1.5.5 Time period ofElectron in nth Bohr's Orbit
^ ^ ^ ^ ^ ^
1.5.6. Current in nth Bohr's Orbit
^
1.5.7 Magnetic Induction at the Nucleus Due to nth Orbit 1.5.8 Magnetic Moment ofthe nth Bohr's Orbit
6 ^
1.5.9 Energy of Electron in nth Orbit
1.5.10 Energies ofDifferent Energy Level in Hydrogenic Atoms
7
1.6 Excitation and Jonization of an Atom 1.6.1 Frequency and Wavelength ofEmitted Radiation 1.6.2 Number ofLines Emitted During, deexcitation ofan Atom
^ 1' 12
1.7 The Hydrogen Spectrum
Chapter? ^
1.7.1 Spectral Series ofHydrogen Atom 1.8 Effect ofMass ofNucleus on Bohr Model 1.9 Use ofBohr Model to Define Hypothetical Atomic Energy Levels
12 21 25
1.10 Atomic Collisions
26
— — — —
DISCUSSION QUESTION CONCEPTUAL MCQs SINGLE OPTION CORRECT NUMERICAL MCQs SINGLE OPTIONS CORRECT ADVANCE MCQs WITH ONE OR MORE OPTIONS CORRECT
33 35 38 42
—
UNSOLVED NUMERICAL PROBLEMS FOR PREPARATION OF NSEP INPhO & IPhO
45
PhotoElectric Effect &MatterWaves
^
.
*^9981
2.1 Electron Emission Processes 2.1.1 Thermionic Emission
2.1.2 Photoelectric Emission
2.1.3 Secondary Emission
30
2.1.4 Field Emission
30
2.2 Photoelectric Effect 2.2.1 Fundamental Laws of Photoelectric Effect 2.3 Experimental Study of Photo Electric Effect 2.3.1 Kinetic Energies of Electrons Reaching Anode 2.3.2 Reversed Potential Across Discharge Tube 2.3.3 Cut off Potential or StoppingPotential 2.3.4 Effect of Change in Frequency of Light on Stopping Potential
2.4 No. of Photon Emitted by Source Per second
31 51
35 36 57 57 58
62
2.5 Intensity of Light due to a Light Source 2.5.1 Photon Flux in a Light Beam 2.5.2 Photon Density in a Light Beam 2.6 Wave Particle Duality 2.6.1 Momentum of a Photon 2.7 DeBroglie's Hypothesis
2.7.1 Explanation of Bohr's Second Postulate 2.8 Radiation Pressure
2.8.1 Force Exerted by a Light Beam, on a Surface 2.8.2 Force Exerted on any Object in the Path of a LightBeam 2.8.3 Force Exerted by a Light Beam at Oblique Incidence
—
Chapters
62 63
63 70 70 70 71 72
72 72
73
2.8.4 Recoiling of an Atom Due to Electron Transition
75
2.8.5 Variation in Wavelength of Emitted Photon with State of Motion of an Atom
75
2.8.6 Variation in Wavelength of Photon During Reflection
75
DISCUSSION QUESTION
80
—
CONCEPTUAL MCQs SINGLE OPTION CORRECT
82
—
NUMERICAL MCQs SINGLE OPTIONS CORRECT
86
—
ADVANCEMCQs WITH ONE OR MORE OPTIONS CORRECT
92
—
UNSOLVED NUMERICAL PROBLEMSFOR PREPARATION OF NSEP, INPhO & IPhO
95
XRays 3.1 Introduction to XRays 3.1.1 Types of Xrays
3.2 Production Mechanism of Xrays 3.2.1 Continuous Xrays
3.2.2 Production of Continuous Xrays 3.2.3 Characteristic Xrays 3.2.4 Production of Characteristic Xrays 3.3 Moseley's Law
991161 100 100
100 100 100 102 102 103
3.4 Applications ofXrays
103
—
DISCUSSION QUESTION
107
—
CONCEPTUAL MCQs SINGLE OPTION CORRECT
108
—
NUMERICAL MCQs SINGLE OPTIONS COI^CT
111
—
ADVANCE MCQs WITH ONE OR MORE OPTIONS CORRECT
113
—
UNSOLVED NUMERICAL PROBLEMS FOR PREPARATION OF NSEP. INPhO & IPhO
115
Nuclear Physics and Radioactivity 4.1 Composition and Structure of The Nucleus
1171861 118
4.1.1 Size of a Nucleus
118
4.1.2 Strong Nuclear Force and Stability of Nucleus
118
4.2 Nuclear Binding Energy
119
4.2.1 Mass Energy Equivalence
124
4.2.2 Binding Energy Per Nucleon
124
4.2.3 Variation of Binding Energy per Nucleon with Mass Number
124
4.3 Radioactivity
127
4.3.1 Measurement of Radioactivity
127
4.3.2 Fundamental Laws of Radioactivity
128
4.3.3 Radioactive Decay Law
128
4.3.4 Half Life Time
129
4.3.5 Alternate form of Decay Equation in terms of Half Life Time
129
4.3.6 Mean Life Time
130
4.3.7 Calculation of Mean Life Time For a Radioactive Element
130
4.4 Radioactive Series
~
135
•
4.4.1 Radioactive Equilibrium
136
4.4.2 Simultaneous Decay Modes of a Radioactive Element 4.4.3 Accumulation of a Radioactive Element in Radioactive Series
136 136 140
4.5 Nuclear Reactions
4.5.1 QValue of Nuclear Reaction
•
•
•
'
'
,
141
141
4.6 Nuclear Fission
4.6.1 Fission of Uranium Isotopes and Chain Reaction
142
4.6.2 Liquid Drop Model
143
4.7 Nuclear Fusion
143
4.8 Properties of Radioactive Radiations
152
4.8.1 Alpha Decay
153
4.8.2 Beta Decay
153
4.8.3 Apparent Violation of Conservation Laws in bdecay
154
4.8.4 Pauli's Neutrino Hypothesis
155
4.8.5 Mass Defect Calculation For bdecay
155
4.8.6 Gamma Decay
156
—
DISCUSSION QUESTION
161
—
CONCEPTUAL MCQs SINGLE OPTION CORRECT
164
—
NUMERICAL MCQs SINGLE OPTIONS CORRECT
168.
—
ADVANCE MCQs WITH ONE OR MORE OPTIONS CORRECT
174
—
UNSOLVED NUMERICAL PROBLEMS FOR PREPARATION OF NSEP, INPhO & IPhO
177
1873401
Geometrical Optics 5.1 Understanding a Light Ray and Light Beams
188
5.1.1 Different Types of Light Rays
188
5.1.2 Different Types of Light Beams
189
5.2 Reflection of Light
189
5.2.1 Regular or Specular Reflection
190
5.2.2 Irregular or Diffused Reflection
190
5.2.3 How we see an object in our surrounding 5.2.4 Laws of Reflection
, '
190
,
5.2.5 Vector Analysis of Laws of Reflection
5.3 Understanding Object and Image in Geometrical Optics
191
191 192
5.3.1 Object in Geometrical Optics
192
5.3.2 Image in Geometrical Optics
193
5.4 Reflection and Image formation by a Plane Mirror
5.5 Field of View for Image formed by a Plane Mirror
194 195
5.5.1 Field of View of an image
195
5.5.2 Field of View of a Mirror for an observer
195
5.6 Characteristics of Image formed by a Plane Mirror 5.6.1 Characteristic} of Image formation by a Plane Mirror
196 196
5.6.2 Characteristic2 of Image formation by a Plane Mirror
196
5.6.3 Characteristic3 of Image formation by a Plane Mirror
197
5.6.4 Characteristic4 of Image formation by a Plane Mirror
198
5.6.5 CharacleristicrS ofImage formation by a Plane Mirror
198
5.6.6 Characteristic6 ofImage formation by a Plane Mirror
198
5.6.7 Characteristic? ofImage formation by a Plane Mirror
198
5.6.8 Characteristics of Image formation by a Plane Mirror
199
5.6.9 Characteristic9 ofImage formation by a Plane Mirror
200
5.6.10 Characteristic10 of Image formation by a Plane Mirror
200
5.6.11 CharacteristicII of Image formation by a Plane Mirror
201
5.7 Understanding Shadow Formation
5.7.1 Umbra and Penumbra Regions 5.7.2 Antumbra Region
201 202 203
5.8 Spherical Mirrors
5.8.1 Standardterms related to Spherical Mirrors 5.8.2 Focal Length of a Spherical Mirror
5.8.3 Image Formation by a Spherical Mirror using Paraxial Rays 5.8.4 Standard Reflected Light Rays for Image Formation by Spherical Mirrors
5.8.5 Relation in focal length and Radius ofCurvature ofa Spherical Mirror 5.8.6 Image formation by Concave Mirrors 5.8.7 Image formation by Convex Mirrors
5.8.8 How an observer sees image ofan extended object in a spherical mirror 5.8.9 How image produced by a spherical mirror can be obtained on a screen 5.8.10 Sign Convention
5.9 Analysis of Image formation by Spherical Mirrors 5.9.1 Mirror Formula for Location of Image
5.9.2 Analyzing Nature ofImage Produced by a Spherical Mirror
5.9.3 Magnification Formula for Size and Orientation ofImage 5.9.4 Relation in Nature and Orientation of Image 5.9.5 Longitudinal Magnification of Image
5.9.6 Superficial Magnification bya Spherical Mirror 5.9.7 Variation Curves ofImage Distance vj Object Distance 5.9.8 Effect ofMoving Object and Spherical Mirror on Image 5.9.9 Effect ofshifting Principal Axis ofa Mirror 5.9.10 Image formation ofdistant Objects by Spherical Mirrors
206
207 207 208 208
209 210 211 212 212 213 214 214 215 215 216 216
217 218 221 223 224 5.9.11 Concept ofReversibility ofLight 224 5.10 Refraction ofLight 226 5.10.1 Absolute Refractive Index ofa Medium 226 5.10.2 Relative Refractive Index ofa Medium 226 5.10.3 Laws ofRefraction 227 5.10.4 Vector form ofSnell's Law ofRefraction 227 5.10.5 Image Formation due to Refraction at a Plane Surface 228 5.10.6 An Object placed in a Denser Medium is seen from Air , 228 5.10.7 An Object placed in Air and seen from a Denser Medium 229 5.10.8 Shift ofimage due to Refraction ofLight by a GlassSlab 231 5.10.9 Shift due to Refraction ofLight by a Hollow thin walled Glass Box placed inside a Denser Medium 232 5.10.10 Lateral Displacement ofLight Ray by a Glass Slab 232 5.10.11 Lateral Displacement ofa Light Ray due to Refraction by Multiple Glass Slabs 5.10.12 Concept ofReflection by a Thick Mirror 5.11 Refraction ofLight by Spherical Surfaces 5.11.1 Analysis ofImage formation by Spherical Surfaces 5.11.2 Lateral Magnification ofImage by Refraction 5.11.3 Longitudinal Magnification of Image 5.11.4 Effect ofmotion ofObject or Refracting Surface on Image 5.12 Total Internal Reflection 5.12.1 Refraction ofLight Rays from a Source in a Denser Medium to Air 5.12.2 Cases ofGrazing Incidence ofLight on a Media Interface . 5.12.3 Refraction by a Transparent Medium of varying Refractive Index
233 233 239 240 243 243 243 247 248 249 249
5.12.4 Total Internal Reflection in a Medium of varying Refractive Index 5.12.5 Equation of Trajectory ofa Light Ray in a Medium of varying Refractive Index 5.13 Prism 5.13.1 Refraction ofLight through a Trihedral Prism • 5.13.2 Deviation Produced by a Small Angled Prism 5.13.3 Maximum Deviation ofLight Ray by a Prism 5.13.4 Condition of a Light Ray to pass through a Prism
250 250 254 255 256 256 257
5.14 Thin Lenses
5.14.1 Converging and Diverging Behaviour ofLenses 5.14.2 Primary and Secondary Focus ofa Lens 5.14.3 Standard Reflected Light Rays for Image Formation by Thin Lenses
264 264 265
5.14.4 Image Formation by Convex Lenses
5.14.5 Image formation by Concave Lenses 5.14.6 Focal length ofa thin lens 5.14.7 Focal length ofdifferent types ofstandard thin lenses 5.15 Analysis ofImage Formation by Thin Lenses 5.15.1 Lateral Magnification in Image Formation by a Thin Lens 5.15.2 Longitudinal Magnification by a Thin Lens
267 268 268 269 269 270
5.15.3 Variation Curves ofImage Distance vs Object Distance for a Thin Lens 5.15.4 Effect ofmotion ofObject and Lens on Image 5.16 Optical Power ofa Thin Lens ora Spherical Mirror 5.16.1 Combination of Thin Lenses 5.16.2 Combination of Thin Lenses and Mirrors 5.16.3 Deviation in a Light Ray due to Refraction through a Thin Lens 5.16.4 Combination of Two Thin Lenses at some Separation 5.16.5 Multiple images produced by a Lens made up ofdifferent materials 5.17 Lens andMirrors submerged in a Transparent Medium 5.18 Displacement Method Experiment to measure focal length ofa Convex Lens
270 271
278
5.19.3 Dispersive Power ofa Prism Material 5.19.4 Dispersion Analysis for a Small Angled Prism
279 280 283 283 284 285 285 286 286 291 292 292 293 293
5.19.5 Achromatic Prism Combination 5.19.6 Direct Vision Prism Combination
294 294
5.18.1 Condition offormation of Real Image by a Thin Convex Lens 5.18.2 Displacement Method Experiment 5.19 Dispersion ofLight 5.19.1 Dispersion of White Light through a Glass Slab 5.19.2 Dispersion of White Light through a Glass Prism
5.20 Optical Aberrations in Lenses and Mirrors 5.20.1 Spherical Aberrations
295 295
5.20.2 Methods to Reduce Spherical Aberrations
295
5.20.3 Chromatic Aberration in a Lens 5.20.4 Achromatic Combination of Lenses
296 297
5.21 Optical Instruments 5.21.1 The Human Eye
201 201
5.21.2 Camera
202
5.21.3 Angular Size ofObjects and Images 5.21.4 Simple Microscope 5.21.5 Magnification ofSimple Microscope 5.21.6 Compound Microscope 5.21.7 Magnifying Power of Compound Microscope 5.21.8 Tube Length of a Compound Microscope 5.21.9 Refracting Astronomical Telescope
202 202 303 203 304 304 204
— — — —
5.21.10 Magnifying Power of a Refracting Telescope 5.21.11 Tube Length of a Refracting Telescope 5.21.12 Reflecting Telescope 5.21.13 Terrestrial Telescope 5.21.14 Galilean Telescope DISCUSSION QUESTION CONCEPTUAL.MCQs SINGLE OPTION CORRECT NUMERICAL MCQs SINGLE OPTIONS CORRECT ADVANCE MCQs WITH ONE OR MORE OPTIONS CORRECT
305 305 306 206 306 311 312 319 325
—
UNSOLVED NUMERICAL PROBLEMS FOR PREPARATION OF NSEP, INPhO & IPhO
330
Wave Optics
^
6.1 Wave Theory 6.1.1 Dual Nature of Light 6.1.2 Wavefront of a Light Wave 6.1.3 Huygen's Wave Theory 6.2 Interference ofLight 6.2.1 Coherent Sources ofLight and Condition of Coherence 6.2.2 Theory ofInterference of Two Waves 6.2.3 Interference of two Coherent Waves of Same Amplitude 6.2.4 Intensity of Light at the Point ofInterference 6.2.5 Condition ofPath Difference for Interference 6.3 Young's Double Slit Experiment (YDSE) 6.3.1 Analysis of Interference Pattern in YDSE 6.3.2 Position of Bright and DarkFringes in YDSE Interference Pattern 6.3.3 Light Intensity on Screen in YDSE Setup 6.3.4 Fringe Width in YDSE Interference Pattern
'
6.4 Modifications in YDSE Setup • ' 6.4.1 Effect of Changing the direction of Incident Light in YDSE 6.4.2 Effect of Submerging YDSE Setup in a Transparent Medium 6.4.3 Path difference between two parallel due to a denser medium inpath of one beam 6.4.4 Effect ofPlacing a Thin Transparent Film infront of one of the slits in YDSE Setup 6.4.5 Concept of zvalue in Interference Pattern of YDSE 6.4.6 Use of White Light in YDSE 6.4.7 Effect of Changing Slit Width in YDSE Setup 6.4.8 Fresnel's Biprism as a Limiting case of YDSE 6.4.9 Lloyd's Mirror as a limiting case of YDSE 6.4.10 Billet SplitLens as a limiting case of YDSE 6.4.11 Interference of Two Converging Coherent Parallel Beams of Light 6.5 Interference by Thin Films 6.5.1 Interference due to Thin Film in Reflected Light at Near Normal Incidence 6.5.2 Interference due to Thin Film in Transmitted Light a Near NormalIncidence 6.5.3 Interference due to a Thin LiquidFilm on Glass 6.5.4 Interference in Reflected Light by a Very Thin Film in Air 6.5.5 Interference in Reflected Lightfrom a Thin Film for Oblique Incidence
6.5.6 6.5.7 6.5.8 6.5.9
Interference in Transmitted Light from a Thin Film for Oblique Incidence Interference in Reflected Light due to Thin Wedge shaped Film Interference by an Air Wedge Shape of Interference Fringes in Reflected Light from different Air Wedges
6.5.10 Shape of Interference Fringes due to different types of Sources
341408; .
^^2 342 342 343 343 344 344 345 345 345 349 349 350 359 351 353 353 354
354 354 355 352 357 35g 359 359 360 3gg 367 368 368 368 368 369 370 37j 371 371
6.6 Diffraction of Light 6.6.1 Explanation of Diffraction by Huygen's Wave Theory 6.6.2 Types of Diffraction of Light 6.6.3 Diffraction of Light by a Single Slit 6.6.4 Analysis of Diffraction of Light by a Single Slit 6.6.5 Diffraction Minima due to Single Slit 6.6.6 Diffraction Minima due to Single Slit
375 375 375 376 377 37g 37g
6.6.7 Observing Single Slit Diffraction Pattern on a Screen
379
6.6.8 Difference between Double Slit Interference and Single Slit Diffraction Patterns 6.6.9 Illumination Pattern due to Diffraction by a Single Slit 6.6.10 Diffraction by a Small Circular Aperture
379 380 3gO
i5.7 Polarization of Light 6.7.1 Representation of Unpolarized and Polarized Light
6.7.2 Circularly and Elliptically Polarized Light
3g2 383 3g4
6.8 'Methods of Polarizing an Ordinary Light
384
6.8.1 Polarization by Reflection
384
6.8.2 Brewster's Law
385
6.8.3 Polarization by Refraction
385
6.8.4 Polarization by Double Refraction
385
6.8.5 Polarization by Dichroism
386
6.8.6 Polarization by Scattering
386
6.8.7 Malus' Law
387
6.8.8 Intensity of Polarized light through a Polaroid (Polarizer)
387
6.8.9 Optical Activity of Substances
388
—
DISCUSSION QUESTION
391
—
CONCEPTUAL MCQs SINGLE OPTION CORRECT
392
—
NUMERICAL MCQs SINGLE OPTIONS CORRECT
395
—
ADVANCE MCQs WITH ONE OR MORE OPTIONS CORRECT
399
—
UNSOLVED NUMERICAL PROBLEMS FOR PREPARATION OF NSEP, INPhO & IPhO
401
answers & SOLUTIONS
Chapter 1
Atomic Physics
409  420
Chapter 2
Photo Electric Effect & Matter Waves
421  434
Chapter 3
XRays
435  440
Chapter4
Nuclear Physics and Radioactivity
441  454
Chapter 5
Geometrical Optics
455  504
Chapter 6
Wave Optics
505  524
1
AtoMic Physics FEW WORDS FOR STUDENTS
,
'
in your previous Classes'from several we have kiidie^d about existence ofatomsand nuclei However most ofus cannotcitemuch experimental evidencefrom them. In this chapter we will discuss
the experiments thatform the basis for our knowledge of atoms: i Nowiweexdmine manyfundamentalfach ofatomic structures and / ourreasonsfor believing inthem. Thus inthis chapter we aregoing 1 tpfranie^ ourgroundworkfor further discussion of atomicphysics ' in the following chapters.
CHAPTER CONTENTS
1.1
hB'riefHistory toAtomic Physics

'"1.6
.1.2 . • Thomson'sAtomic Model • >; r '1.3
Ruthorford'sAtomicModel
1.4.
Bohr's Modelofan Atom
1.5
•Excitation andlonization ofan Atom
1.7
TheHydrogen Spectrum
.
'Effect.ofMass ofNucleusonBohrModel
 1.8
1.9
~
, Use ofBohrModH toDefine Hypothetical Atomic .
Properties ofElectron in Bohr's AtomicModel
•'
Energy Levels.
, .
,
1.10 ' Atomic Collisions 
^ii:i
COVER APPLICATION
•
Goldfol
Alpha Particles
Alpha partilces Source
An Atom
of Go d
. iLrad
Detector
Figure(a)
i'.
•
'
. '
Figure(b)
Ruthorford's Alpha Scattering experiment based on which atomic model was defined by Ruthorford. Figure(a) shows.the experimental setup and figure(b) shows the atomic visualization of scattering of alpha particles.
'
"
• •:
Atomic Physics is that branch of physics in which we'll deal with the properties ofan atom. Thewhole Physics is based on ourphysical environment, surroundings andeverything inour surroundings is the multidimensional arrangement ofdifferent atoms.
.Afeniic Ph^}t%j
The actual observations are quite different. At rooiti temperature
or belownormalroom,hydrogenis vary stable,it neither emits radiation nor does the electron collapse into the nucleus. The
Hydrogen atom emits fixed wavelengths which areobserved in Hydrogen spectrum. Such observations could notbeexplained by classical concepts of physics.
Therewere somanyph)«icists whohad given theirexplanation
Lenard and Rutherford but no one was successful with his
These puzzles related to the Hydrogen atom andits spectrum was solved upto a greater extent byanatomic model which was presented byNiels Bohr, in 1913. Thiswas theageofModem
theorywhichcouldmakean atoma femiliarconcept.
Physics. "
Initiallythe wholeresearch was concentrated on the simplest atomHydrogen Atom andits spectrum. When a material body is heated, it emitselectromagnetic radiation. The radiationmay consists ofvarious components havingdifferent wavelengths. When these wavelengths are plotted on a calibratedscale then this plot is called the spectrumofthe material body.
Now we'll discuss this ^Bohr's Model ofatom" in detail with
If hydrogen gas enclosed in a sealed tube is heated to high temperatures, it emits radiation. If this radiation is passed through a prism, components of different wavelengths are deviated by different amounts and thus we get the hydrogen spectrum. When the'Hydrogen spectrum was taken then the mainfeatures ofthis spectrum are,somesharplydefined, discrete wavelengths exist in the emitted radiation. For example, the radiation of wavelength 6562 A was observed and then the radiation of4860 A was observed. Hydrogen atom do not emit
1.2 Thomson's Atomic Model
about the nature of an atom with their atomic models and
suggestions. Main physicists in this field were Thomson,
any radiation between 6562 Aand 4860 A. There were also so manylines ofelectromagnetic radiations above and below these
theoretical andanalytical expressions. Theatomic model of Bohr is based on three assumptions, these assumptions are called Postulates of the Bohr's model. Before proceeding to Bohr's Model First we'll discuss the basic concepts of Thomson's Model and Rutherford's experiment.
In 1898, J.J.Thomson suggested an atomic modelwhichbecame the most popular model of an atom in very early twentieth century. He suggested that the positive atom was spherical in
shape and the positive charges were uniformly distributed in thissphere of atomic dimensions orthe sphere was filled with positive charged matter ofuniform density andsufficient number of negative particles, electrons are embedded randomly throughoutthe volume just like seedsin watermelon to balance the positive charge.
two lines in visible region, ultraviolet, infia red region etc.
1.1 A Brief History to Atomic Physics In this age of classical Physics no one was presentwho could explain the theoryofhydrogen spectrum. Although Rutherford got someinteresting resultswith his ascattering experiment
Uniformly distributed positive charge
and based on these observations, Rutherford proposed the model ofnuclear atom which remain accepted to a large extent
even today. According to this model all the positive chargeof the atom is concentrated at its centre called the nucleus of the atom. This nucleus contains almost all the mass of the atom. Outside this nucleus, there are electrons which move around it
Figure 1.1
at somesqjaration. This modelwasnot fullysatisfactorybecause it was not able to explain the contradictions,imposed by the classical theory of electromagnetics. According to Maxwell's electromagneticequation any acceleratedcharged particle must continuously emit electromagnetic radiation. The revolving electron should therefore, always emit radiations at all
Figure1.1 shows an approximate,picture of an atom what Thomson suggested.'Here the gray coloured sphere can be assumedas the positivelycharged uniformmatter and electrons (representedas small balls) are randomlyscatteredwithin the
temperatures and the radius of the circle should gradually
positive charge resembled the plum fiuit seeds in a pudding
decrease and the electron should finally fall into the nucleus. But this is not the case actually exist.
hence this atomic model is sometimes called as "plum pudding
volume of this sphere. Here the electrons placed within the
moder.
lAtomic Physics _
1,3 Ruthorford's Atomic Model
In theRuthorford's planetarymodel, twobasicdifficulties exist.
In 1911, Earnest Rutherford performed acritical experiment that
electromagnetic radiation by an atom. Itisobserved that every
First is the emission of some characteristic frequencies of
showed that Thomson's model could not be correct. In this
atom emits some characteristic frequencies and no other
experiment a beam ofpositively charged alpha particles (helium
frequency isemitted. Ruthorford's model was not able toexplain
nuclei) was projected into a thin gold foil. It is observed that most of the alpha particlespassed through the foil as if it were emptyspace. But somesurprising results are also seen. Several
this phenomenon. Second difficulty was the revolution of electrons around the nucleus. These electrons undergo a centripetal acceleration. According to Maxwell's Theory of alpha particles are deflected from their original direction by electromagnetism, a centripetally accelerated, charge particle large angles.'Few alpha particles are observed to be reflected should continuously radiate electromagnetic waves of same back, reversing their direction oftravel as shown infigure1.2. frequency asthat ofitsrevolution. Inthis model ifwe apply this classical theory it says that as electron radiates energy, the Viewing radius of its orbit steadily decreases and its frequency of screen
revolution continuously increases. This would lead to an ever
increasing frequency ofemitted radiation andultimately atom will collapseas the electronplunges into the nucleus as shown in figure1.4.
Source screen
Figure 1.2
IfThomson model isassumed to betruethatthe positive charge is spreaded uniformly in the volume of an atom then the alpha particle can never experience such a large repulsion due to Figure 1.4 which it will be deflected by such large angles as observed in the experiment. On the basis of this experiment Ruthorford 1.4 Bohr's Model of an Atom
presented a new atomic model.
Between 1913 and 1915 Niels Bohr developed a quantitative In this new atomic model
atomic model for the Hydrogen atom that could account for its spectrum. The model incorporated the nuclear model of the
it was assumed that the
positive charge in the atom was concentrated in a region that was small
/ /
relative to the size ofatom.
'
He
1
called ,
this
atom proposedbyRuthorfordon the basis of his experiments. We shall see that this model was successful in its ability to predictthe gross features of the spectrum emittedbyHydrogen atom. This model was developed specifically for Hydrogenic
\
concentration of positive ' \ charge, thenucleus ofthe
'
atoms. Hydrogenic atoms are those which consist ofa nucleus
\
f
j /
atom. Electrons belonging
interactions in an atom are not accounted in the Bohr's Model
JBOVi "to the atom were assumed'
that's why it'was valid only for one electron system or
to be moving in the large ^
volume of atom outside
with positive charge + Ze (Z = atomic number, e = charge of electron) and a singleelectron. Morecomplex electronelectron
hydrogenic atoms. Figure 1.3
the nucleus. To explain why these electrons were not pulled
The Bohr model is appropriate for one electron systems like H,
into the nucleus, Ruthorford said that electrons revolve around
He"^, Li"^^ etc. and itwas successful upto some extent inexplaining
the nucleus in orbits around the positively charged nucleus in the same manner as the planets orbitthe sun. Thecorresponding
the features of the spectrum emitted bysuch hydrogenic atoms. However this model is not giving a true picture of even these simple atoms. The true picture is fiillya quantum mechanical
atomic model can be approximately shown in figure i .3.
Atomic physics
. affair which is different from Bohr model in several fundamental
1.4.2 Second Postulate
ways. Since Bohr model incorporates aspects ofsome classical
and somemodemphysics, it is nowcalledsemiclassical model. In the studyof atom, Bohr found that while revolving around Bohr has explained his atomic model in three steps called the nucleus the orbital angular momentum of the electron was . postulates ofBohr's atomic model. Lets discuss thesepostulates restrictedto onlycertain values, we saythat the orbitalangular momentum ofthe electron is quantized. He therefore took this one by one. as a second postulate of the model. The statement of second Postulate is, Bohr proposed that  ''During revolution around 1.4.1 First Postulate the nucleus, the orbital angular momentum ofelectronL could In this postulate Bohrincorporates andanalyze features ofthe not havejust any value, it can take uponlythosevalueswhich Ruthorford nuclear model ofatom. In this postulate it was taken are integral multiples of Plank's Constant divided by 2it that as the mass ofnucleus is so much greater then the mass of electron, nucleus was assumed to be at rest and electron
h .. I.e.
271
revolves around the nucleus in an orbit. The orbit of electron is
assumedto be circular for simplicity.Now the statement of first postulateis ''^During revolutionofelectron around the nucleus
Thus the angular momentumof electroncan be written as T ~
—
in circular orbit, the electric coulombianforce on electron is
balanced by the centrifugalforce acting on it in the rotating frame ofreference"
In
...(1.4)
Where«is a positiveinteger, knownas quantumnumber. In an orbit of radius r if an electron(mass m)revolvesat speedv, then its angular momentum can be given as L = mvr
...(1.5)
Now from equation(1.4) and (1.5), we have for a revolving electron nh mvr =
In
...(1.6)
Equation(1.6) is known as equation of second postulate of
Bohr model. Here the quantity ^h occurs so frequently in
Figure 1.5
If electron revolves with speed v in the orbit ofradius r. Then relative to rotating frame attached with electron, the centrifugal force acting on it is
modern physics that, for convenience, it is given its own designation fi, pronounced as "Abar." In
mv
F r= cf
r
.
 1.055 xlO^'^Js
...(1.7)
...(l.I) 1.4.3 Third Postulate
The coulombian force acting on electron due to charge of While revolution of an electron in an orbit its total energy is taken as sum of its kinetic and electric potential energy due to the interaction with nucleus. Potential energy of electron
nucleus (+ Ze) is F
electric
K{e){Ze) ^2
revolving in an orbit of radius r can be simply given as KZe'
...(1.2)
[/=
K{e){Ze)
r~
Now according to first postulate from equation( 1.1) & (1.2) we have
.
U=
mv^
mv^ =
KZe^
KZe'
KZe'
...(1.8)
For kineticenergyofelectron,we assumethat relativisticspeeds
...(1.3)
Equation(l .3) is called equation ofBohr's flrstpostulate.
are not involved so we can use the classical expression for kinetic energy. Thus kinetic energy of electron in an orbit revolving at speed v can be given as
K=\mv'
...(1.9)
iAtomic^PKysics,'  '
' _5;:
Thus total energy ofelectron can begiven as KZe'
E = K+ U=
1.5 Properties of Electron in Bohr's Atomic Model ...(1.10)
Here we can see that while revolving inastable orbit, the energy of electron remains constant. From the purely classical viewpoint, during circular motion, aselectron isacceloated, it should steadily loose energy by emitting electromagnetic
Now we'll discuss the basic properties ofan electron revolving in stable orbits. These stable orbits we call Bohr energy level. We have discussed thatthere aresome particular orbits inwhich electron can revolve around the nucleus for which first and
second postulates ofBohr model was satisfied. Thus only those orbits are stable for which the quantum number «= 1,2,3
radiations and it spiraled down into the nucleus and collapse Now for n^ orbit ifwe assume its radius is denoted by r and electron is revolving in this orbit with speed v^. We can represent
the atom.
Bohr in histhirdpostulate stated that''While revolving around
the nucleus inanorbit, itisinstable state, itdoes not emit any
all the physical parameters associated with the electron in n^'
orbit by using a subscript n with the symbol ofthe physical
parameter like r^, v^ etc. ener^ radiation during revolution. It emits energy radiation only when it makes a transition from higher energy level 1,5.1 Radius ofn^''Orbit in Bohr Model (upper orbit) to a lower energy level (lower orbit) and the energ;ofemittedradiation isequal tothe difference inenergies Radius ofelectron in Bohr's Orbit can be calculated using of electron in the two corresponding orbits in transition.'' thefirst two postulates ofthe Bohr's model, using equations(1.3) & (1.6) wehavefrom equation(1.6)
Ifan electron makes atransition form ahigher orbit n.^ to alower orbit w, as shown in figure1.69 (b). Then theelectron radiates
nh V
a single photon of energy
=
"
2nmr„
Substituting this value of inequation(l .3), weget
A£ =£„2
=hv
2.2
Wh
r
...(1.12)
An^KZe^m r_ =
An^Ke^m
=0.529 xA (a)
...(1.13)
(b)
Figure 1.6
Here
and E„^ are the total energies ofelectron inthe two
1.5.2 Velocity of Electron in
Bohr's Orbit
orbits «2 and n^ The emitted photon energy can beexpressed
By substituting the value of r^ from e9uation(1.12) in
as hv where v is the frequency of radiated energyphoton. If be the wavelength ofphotonemittedthen the energyofemitted
equation( 1.6) we can calculate the value ofv as
photon can also be given as
InKZe'
...(1.14)
V
nh
/^ =hv=^ K
...(1.11)
Similarlywhen energyis supplied to the atom by an external source thentheelectron will make atransition from lower energy level to a higher energylevel as shown in figure1.6 (a). This process is called excitation of electron from lower to higher energylevel. In thisprocess the wayinwhich energyis supplied to the electron is very important because the behaviour of the
2nKe^ V. =
h
v^ =2.18 Xlo^x —m/s 1.5.3 Angular Velocityof Electron in
...(1.15)
Bohr's Orbit
Angular velocity of the electron in n"" orbit is given by
electron in the excitation depends on the process by which energy is supplied from an external source. This we'll discuss in
CO. =
detail in'later part of this chapter.
First we'll studythe basicproperties of an electron revolving around the nucleus of hydrogenic atoms.
CO.. =
...(1.16)
• .
^
1.5.4 FrequencyofEIectroniii«*^Bohr'sOrbit
M^ =/^x7ir^
Frequency i.e. the number ofrevolution made by the electron j. th . per second in« orbit IS given as
M," "
 j 'Atorriic,..Ph^ps1 •
r~5 n h
^^
^An^KZe^m
CO
^
y=
KZ e m
inthisequation(1.21), for «= Ithe magnetic moment M, is also
"
known as"5o^rMjg/2ero«".
1.5.5 Time period of Electron in
Timeperiodof electron in
Bohr's Orbit
1.5.9 Energy ofElectron in
orbitis given by
Orbit
We've discussed that in orbit during revolution the total energyof electroncan be given as sum of kinetic and potential
rn =4
energy of the electron as
r
nm
E =K+U
...(1.22)
An^K^Z^e^m Kinetic energy ofelectron in 1.5.6. Current in
orbit can be given as
Bohr's Orbit
...(1.23) Electrons revolve around the nucleus in the «
Bohr's Orbit
then due torevolution there iscurrent inthe orbit and according to the definition of current, the current in the
From equation offirst postulate ofBohr Model we have for
orbit willbe orbit
total coulombs passing through a point in one seconds, and in
an orbit an electron passes through a pointy^ times in one second so the current in the w'** orbit will be
=>
O 1 T
'•
E =
In^K^Z^e^m :r:
"
n 2f'2'72 4_
„
=>
E=Rch joules and iscalled as One RydbergEnergy
^n~~
In K
Z
e
m
1Rydberg= 13.6eV=2.17x 10''^joules
The equation(1.29) clearly shows that as the value of« j«ubetween ^ increases, *1. the difference two consecutive energy 11 levels
decreases. It can be shown with the help of Figure1.7, which shows the energy level diagram for a hydrogen atom.
2nr„ .
2.18xl0^x%
/« =
T7 s"
2x3.14x0.529x10"'° x"/2
. b Thusnuraberofrevolutionscompletedm 10 ®secondin« =2 ^^ . state are
A^=/^x 10"^
—^0 n = 6 —
«= 5
2.18x10^x10"^
A rr rPsnsdpV z — 2 x^3 . 1 4 x 0 .529x10"'°
^
„=4
£=0.8SeV
n =3
1.51 eV
„ =2
E—ZAeV
^
X ——
3 2 x 10^revolutions
# Illustrative Example 1.2
£ = 13.6 eV
« = l
Figure 1.7
What is the angular momentum ofan electron in Bohr's hydrogen atom whose energy is  3.4 eV ?"
Now if we multiplythe numerator and denominatorofequation(1.28)by c/iwe get Solution X
ch
=>

Where R=
E^=Kchx^eVn
^
X —^
Energy of electron m n Bohr orbit ofhydrogen atom is given " . , by,
n
,
...(1.30)
E=~^eV ,n
defined as Rydberg Constant and jjence
34^ ^3.6
thevalue ofit isgiven as/?= 10967800 m"',which can betaken
"
approximatelyas lO'm"'.For«= 1andZ= 1theenergyisgiven
^
« =4
as
=>
«=2
•
_
Atomic Physics^
The angular momentum of an electron in
orbit is given as
# Illustrative Example 1.5
yih
L= X. Putting « = 2, we obtain
An electron in the ground state ofhydrogen atom is revolving
iTZ
in anticlockwise direction in the circular orbit of radius R as
r^2A=:h. 2%
shown in figure1.8. (i) Obtain an expression for the orbital magnetic dipolemoment
71 '
ofthe electron.
# Illustrative Example 1.3
The electron in a hydrogen atom makes a transition Wj wheren, and arethe principlequantum numbers ofthe two states. Assume the Bohr model to be Valid, the time perioil of the electron in the initial state is eight times that in the final
state. What are the possiblevalues of
and M2 ? Figure 1.8
Solution
The time period T of an electron in a Bohr orbit of principal quantum number n is n'h
Solution
Tccn^
(i) According to Bohr's second postulate
li T,
with the magnetic induction. Find the torque experienced by the orbiting electron.
3.3
Thus,
(ii) The atom is placedin a uniform magnetic inductionB such that the plane normal ofthe electron orbit makes an angle 3.0°
4
h
h
mvr= n — =
—
2n
As Tj = STj, the above relation gives'
2k
' (As for n = 1 first only) •
«2 j =>

h'
•
'
M] = 2^2
...(1.31)
2nmr^
Weknow that the rate of flow of charge is current. Hence current
Thus the possible values of Wj and Uj are
in first orbit is
nj=2,«2=l; n] = 4, «2 = 2;
Now fi^om equation(l .31) i = ev,= e
= 6, rt2~3; and so on # Illustrative Example 1.4 =>
How many times does the electron go round the first Bohr orbit ofhydrogen atom in b ?
1 =
2717*1
2717*1
27cr]
XVi
h
eh
2Kmr^
4^^^ m r 2
...(1.32)
Magnetic dipole moment,
Mj = i X Solution M,=
We know the fi^equencyof revolution of electron is
• eh
eh
An'^mr^
4Km
...(1.33)
(li) Torque on the orbiting electron in unifonn magnetic field is it is the number ofrevolutions in 1 second, it can be given as for
X = M
X B
n = 1 orbit of hydrogen atom as
/i =
T = Afflsin30°
2.18x10^ 2x3.14x0.529x10
,15 /j = 6.56 X10'= sec
10
1 sec
eh x=
AKtn
B x  ^ =•
'2
ehB %Km

# IllustrativeExample 1.6
Practice Exercise 1.1
Determine the maximum wavelength that hydrogen in its ground state can absorb. What would be the next smaller wavelength that would work ? Solution
(i)
The innermost orbit ofthe hydrogen atom has a diameter
of 1.06A. Whatis the diameter of the tenthorbit? • [106 A]
. . .
•
. '
(ii) Which energystate ofthe triplyionized beryllium (Be^ has the same electron orbital radius as that ofthe ground state
Maximum wavelength will correspond to the minimum energy ofhydrogen? Given 2 for beryllium =4 : transition ofanelectron. From ground state ofhydrogen atom
the minimum energy transition is forn = Lton = 2, for which energyreleased willbe

[« = 2]
(iii)
•
^
InQ. (ii), what isthe ratio'ofthe enef^ state ofberyllitim
and that of hydrogen ? ri
'
[4]
• A£:,2 = (3.4eV)('13.6eV)
(iv) The orbital speed oftheelectron in the ground state of hydrogen isv. What will beitsorbital speed when it isexcited to
A5,2= 10.2eV
Thus 10:2 eV energy isabsorbed inthe form ofaphoton, ifX. be the energy state3.4 eV ? its wavelength then

•: [^l
he
Y =10.2 X16 xi(r'9 x=
^
..... (v) Which energy state ofdoubly ionized lithium (Li^ has the same energy as that of the ground state of hydrogen ?
he
10.2x1.6 xlO~'^
Given Zfor lithium = 3. [« = 3]
,.(6,63x10'^'^)x(3x1Q:'^) 10.2x1.6x10:^^ ,
• A=12I8A
(vQ . ^In the Bohrmodel ofthe hydrogen atom, find theratioof,
•• ( .
the kinetic energy to Aetotal energy ofthe electron inaquantum ;
Fornext smaller wavelength the possibility is for an electron
transition from « ^ l;ton = 3, for which the absorbed energy photon requiredds
j
state«?
, . .
,,
[1]
(vii)
. ^
,,
•
The total energyofthe electronin the first excitedstate
ofhydrogen is 3.4 eV. What isthekinetic energy ofthe electron
ae,,=e,e,
in this state?
A£i3 =(:i.5kV)(13.6eV)
,
,
[+ 3.4 eV]
:'^A£,3 = 12.09ey
If X'be itswavelength then "it is given as
'•
1.6 Excitation and lonizatioh of an Atom'
he
• '7^= 12.09 eV
'
•
(6.63xlO"^'')x(3xlO'*)' ^ ' ^ 12.09x1.6x10"'^ X' = 1027.5A
According to third postulate of Bohr model we've discussed when some energy is given to an electron of atom from an
external source it niay make a transition to theupper energy level.This phenomenon we call excitation ofelectronor atom and the upper energy level to which the electron is excited is
called excited state. To excite an electron toa higherstate energy
can besupplied to it in two ways. Here we'll discuss only the Web Reference at.ww.phvsicsgalaxv.com
Age Group High School Physics j Age 1719 Years
energy supply by an electromagnetic photon. Other method of .energy supplywe'll discusslater in this chapter.
SectionMODERN PHYSICS
Topic  Atomic Structure Module Number  r to 18
According to Plank's quantum theoryphoton is.defined as a packet of electromagnetic energy, which when absorbed bya
physical particle, its complete electromagnetic energy is
Atomic Physics
10
converted into the mechanical energy of particle or the particle
;£., = OeV
utilizes the energy of photon in the form of increment in its mechanical energy. When a photon is supplied to an atom and an electron absorbs this photon, then the electron gets excited to a higher energy level only if the photon energy is equal to the difference in energies ofthe twoenergy levels involved in
;£4 = 0.85 eV .£3 = 1.51 eV A£,3= 12.09eV
E
—2.6 eV (Hvpothctical level)
•£, =•3.4 eV A£,2=102eV
=
eV
£, = 13.6eV
the transition. Figure 1.9
For example say in hydrogen atom an electron is in ground
state (energy E^~ 13.6 eV). Now it absorbs a photon and As discussed in previoussections, in an atom electroncan not makes atransition to«=3 state (Energy£3=1.51 eV) then the takeup all energies. It can existonlyin someparticular energy levels which have energy given as 13.6/«^e V. energy of incident photon must be equal to A£,3 = £3£, M
,3 =(l51)(I3.6)eV
A£,3 = 12.09 eV Similarlyif wefind the difference in energiesof staten = 1 and w= 4, we get
^14 ^4 A£,4= (0.85)(13.6)eV
A£,4= 12.75 eV Thuswhen a photon ofenergy 12.75 eVincidaitsonahydrogen
atom, the electron maybe excited to«=4 level ifit absorbs this photon. In the same fashion, the energy ofa photon required to excite the electron ofa hydrogenatom from ground state (w = 1)
to next higher level {n = 2) which we call first excited state is given as
Whena photon of energy 11 eV is absorbed by an electron in groundstate. The energy ofelectron becomes  2.6eVor it will excite to a hypothetical energy level X somewhere between n = 2 and « = 3 as shown in figure1.9,which is not permissible for an electron. Thus when in ground state electron can absorb
onlythosephotons whichhaveenergies equal to the difference in energies of the stable energy levels with groundstate. If a photon beam incidenton Hatoms having photon energynot equal to the difference of energy levels of Hatoms such as 11 eV, the beam willjust betransmittedwithoutany absorption by the Hatoms.
Thus to excite an electron fiom lower energy level to higher
levels by photons, it is necessary that the photon must be of energy equal to the difference in energies of the two energy levels involved in the transition.
As we know that for higher energy levels, energy of electrons is less. When an electron is moved away from the nucleus to
00* energy level or at « = 00, the energy becomes 0 or the A£i2 =(3.4)(13.6)eV
electron becomes fiee fi*om the attraction of nucleus or it is removed fiom the atom. Infect when an electron is in an atom,
A£,2 = 10.2eV its total energy is negative
Here we have seen that in the ground state of a hydrogen atom electron can absorbs photons ofenergies 10.2eV, 12.09eVand 12.75eV to get excitedto « = 2,« = 3 and « = 4 levels.
E„ =~1Mz2
. This negative
sign shows that electron is under the influence of attractive
Now we'll see what will happen when a photon of energy equal
forces of nucleus. Whenenergyequal in magnitudeto the total energy of an electron in a particular energy level is given externally, its total energy becomes zero or we can say that energy, level or the electron is electron gets excited to
to 11 eV incident on this atom. From the above calculation of
removed fiom the atom and atom is said to be ionized.
energy differences of different energy levels we can say that if the electron in ground state absorbs this photon it will jump to a state somewhere between energy levels n = l and w = 3 as shown in figure1.9. When electron in ground state absorbs a photon of 11 eV energy, its total energy becomes
£ = £1 + 11
We know that removal of electron fiom an atom is called
ionization. In other words, ionization is the excitation of an electron to « = 00 level.The energy required to ionize an atom is called ionization energy of atom for the particular energy level fiom which the electron is removed. In hydrogenic atoms, the
ionization energy for =>
£=13.6 + lleV
=>
£=2.6eV
A£.
state can be given as
= £„£_
lAitoffiiciiPhysJcs A£"
11
13.6Z
=0
This energy can.be converted toeVhy dividing this energy by theelectronic charge e,asifwavelength isgiven in the energy
2^ eV
n
in eF can be given as 13.62^
A£=^(ineV)
eV
...(1.36)
Substituting the values of h, a and e weget
1.6.1 Frequencyand Wavelength ofEmitted Radiation
(6.63xl0^'')x(3xl0^)
When an electron absorbs a monochromatic radiation from an external endrgy source then it makes a transition fromTa lower energy level to a higher level. But this state of the electron is not a stable one. Electron can remain in this excited state for a
verysmall internal at most of the order of 10"^ second.The time period for whichthis excited state ofthe electronexists is called
Xx(1.6xl0^l^)xl0"^° ^
AE=^^eV
...(1.37)
Hereinaboveequation(1.37), lambdais in A units.
the life time ofthat excited state. After the life time ofthe excited
state the electron mustradiate energy and it willjumpto the Thisequation is themost important in numerical calculations, ground state. Sayif it wasin fifth energy level thenit maycome asitwill bevery frequently used. From equation(1.34) &(1.36) totheground state byfollowing the path as5 ^ 3 ^ 1or it may we have follow 5>4>2^Iorin somanyways, andin eachtransition it willemita photon ofenergyequaltothe energy difference of the two corresponding orbits according to Bohr's Third
1
1
«1 V"i
n
eV
Postulate. 13.62
Let us assume that the electron isinitially in «2 state and itwill
he
1 Wi
jump to a lower state «, then it will emita photon of energy
equal tothe energy difference ofthe two states «, and
1 n
eV
2J
as
v=^=RZ^
1
1
w,
n.
...(1.38)
Where AE" isthe energy ofthe emitted photon. Nowsubstituting
the values ofE„^ ax\.&E„^ in above equation, we get AE = 
Here v is called wave number of the emitted radiation and is
+ 21.2 nth
'~t—2 ir2 3 ^ 2 ^ 1.Thusmaximum number of photons"emitted bya single electroh from state « are A^=«l
It is clear that the energy of outer orbits is greater than the energy of inner orbits. When external radiation is given to the hydrogen atom then the electron in ground state jumps to a higher energy state and the atom is called now in excited state.
Anyexcited stateis an unstable stateand themaximumlifetime ofan excited state is ofthe orderof 10"® seconds, and afterthe lifetime of the excited state the electron jumps to the ground state again directly or indirectly by emitting one or more electromagnetic radiations. It may have so many paths to come to ground state.
,• .,
,
The wavelength of the lines of every spectral series can be calculatedusing the formulagivenbyequation( 1.37).. _
Five spectral sbies are observed in the Hydrogen Spectrum correspondingto the five energy levels of the Hydrogen atom and these five series, are named as on the names of their inventors. These series are
(1) Lyman Series (3) Paschen Series ...(1.41) '(5) Pfiind Series
1.7 The Hydrogen Spectrum
.,j
(2) Balmer Series (4) Breckett Series
These spectral series are shown in figure1.11 n =
£=0
6.
« = 7
h = 6
£=0.54 eV. n = 5
Pfund Series
n = 4
£ = 0.85 eV
Brackett Scries £ =  1.51 eV
II = 3
Paschen Series £ = 3.4 eV
« = 2
Balmer Series £. = 13.6 eV
«= 1
Lyman Series Figure 1.11
When Hydrogen gas is discharged in a discharge tube (at High
(1) Lyman Series : The series consists of wavelengths of the
potential difference of the order of 10'^ volts), the Hydrogen
radiations which are emitted when electron jumps from a higher
iAjgnicfPh^ics
13
energy level to «= 1orbit. The wavelengths constituting this series lie in the Ultra Violet region ofthe electromagnetic spectrum,; ;
1
1
...(1.42)
, . •
For the first member of Lymanseries For Lyman Series v= R
•«i = l and
1
I
M
...(1.43)
4
n.=2,3,4.
Dividing equation(l .42) byequation(l .43), we get
First line of Lyman series is the line corresponding to the transition «2 ~ 2 to «[ = 1, similarly second line ofthe Lyman
^ = A X.
97 27
series is the line corresponding to the transition =3to Wj =1.
^ T  At,
5x6563
^
07 27 ^ 1
= 1215.37 A.
27 . (2) BalmerSeries : The series consists ofwavelengths pfthe radiations which are emitted when electronjumps from ahigher energy level to w= 2 orbit. The wavelengths consisting this # Illustrative Example 1.8 series liein thevisible region oftheelectromagnetic spectrum. Find the ratio ofionization energy ofBohr's hydrogen atom and (3) Paschen Series :The series consists of wavelengths of hydrogenlike lithium atom. the.radiations which areemitted when electron jumps from a higher energy level to«= 3 orbit.The wavelengths constituting Solution this series lieinthe Near Infra Red region ofthe electromagnetic Energy of an electron in ground state ofBohr's hydrogenlike spectrum.
atom is given by,
(4) Brackett Series : Theseries consists of wavelengths of the radiations which areemitted when electron jumps from a higher energy level to«=4 orbit. The wavelengths constituting this series lie in the Infra Red region of the electromagnetic
E=
13.6Z
eV
WhereZ=atomicnumber oftheatom.
spectrum.
I
The ionization energy of this atom is equal in magnitude to (5) Pfund Series ; The series consists of wavelengths of the radiations which areemitted when electron jumps from ahigher energy level to « = 5 orbit. Thewavelengths constituting this series lie in the Deep InfraRed region ofthe electromagnetic spectrum. >: 
energy ofground state =E^= 13.6 Z^.
(EJn
{Zuf
(EJm _(l
l3
Wecan findoutthe wavelengths corresponding to the firstline and the last line forremaining four spectralseries as mentioned in the case ofLyman Series.
a Illustrative Sample 1.9
Lets discuss the transition ofelectron between different orbits
Electronsofenergies 10.20eV and 12.09eV can causeradiation
indetail with thehelp ofsome examples.
to beemitted from hydrogen atoms; Calculate in eachcase, the principal quantum number of the orbit to which electron in the
# Illustrative Example 1.7^
hydrogen atom is raised and the wavelength of the radiation emitted if it drops back to the ground state.
The wavelength of the first member of the Balmer series in
hydrogen spectrum is 6563 A. Find the wavelength of first
Solution
memberof Lyman series in the samespectrum. Solution
For the first member ofthe Balmer series
'
We know the orbital energy ofan electron revolving in
orbit
is given by
•
r 13.6 E=^eV
.14
Where n is the principal quantum number.
Solution
When
We know for transition w= 2 to « = 1, the energy released is
n=l E, =13.6eV
« = 2£2=3.4eV n = 3£31.51eV
=>
A£2, =(3.4eV)(13.6 eV)
Here we can see that
The wavelength corresponding to this transition can be given 10.0eV = £2^i and
as
\2.09ey = E^E^
12431 X=
Thus by absorbinga radiation photon of 10.2 eV electronwill
10.2
= 1218.7 A
This radiations is in ultraviolet region.
make a transition to « = 2 state and by absorbing 12.09 eV
photon electron willmakea transition to « = 3 state. Nowafter # Illustrative Example 1.11 the life time ofexcited states, the electron in w= 2 and n = 3 will make transitions to lower states and ultimately come back to
ground state. In this process the possibilities of reverse transition are
n = 3
to
n= 2
n = 3
to
«= 1
« = 2
to
w= 1
In above three transitions the amount of energy released will be
A£32 = (1.5IeV)(3.4eV)
Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 975 A. How many different linesare possible in theresulting spectrum? Calculate the longest wavelength amongst them. You may assume the ionization energyfor hydrogen atom as 13.6 eV. Solution
When an electron of hydrogen atom absorbs a wavelength
975 A. The energy in eV, it absorbs is given by •
= I.89eV
A£3, =(1.51 eV)C I3.6eV) = 12.09 eV
A^2i =(3.4eV)(13.6 eV)
=>
^
975
£ = 12.75 eV
As discussed earlier that 12.75 eV is the energy difference of n  l.and nA. Thus electronwill make a transition from ground state to w= 4 orbit. From n = A when electron will start reverse
= 10.2eV
Thus wavelengths of radiations of corresponding transition
transitions and ultimately comesback to ground state then the number of possible transitions are
are
32
1.89
6577.2 A XT N= —^ =6,
^31
_ 12431 12.09 = 1028.2 A
,
_ 12431
These are also shown in figure1.12 £.,=0.85eV
n = 4
A,.,i —
21
10.2
= 1218.7A
i
« = 3
•£3 —1.51 eV •£2=3.4 eV
n = 2
# Illustrative Example 1.10 M= 1
Determine the wavelength ofthe first Lyman line, the transition from«=2 to « = 1.In what region ofthe electromagneticspectrum does this line lie ?
£i=13.6eV Ground State
Figure 1.12
omTc^ Physics
•Among these six transitions, the longest wavelength will be corresponding to the minimum energy which is released for transition «=4 to« = 3, theenergyreleased in thistransition is
(ii) Energy level diagram isshown infigure1.13. n = 4 (0.85 eV) « = 3 (1.51 eV)
^^43 ^4 ^3
=> =>
n = 2 (3.4eV)
 A£'43 = (0.85eV)(~1.51eV) A£43=0.66eV
n=l (13.6eV)
The wavelength' correspondingto this energy is Figure 1.13 12431 0.66
A
X= 18834.8 A
Illustrative Example 1.12
A hydrogenatom in a statehaving a bindingenergyof 0.85 eV makes a transition toa statewith excitation energy 10:2 eV. (i) ^•Find the energyand wavelength of the photon emitted. (ii) Show the transition on an energy level diagram for hydrogen, indicating quantum numbers.
# Illustrative Example 1.13
Ultraviolet lightofwavelength 830A and 700A when allowed to fallon hydrogen atomin their ground stateis found to liberate
electrons with kinetic energy 1.8 e'V and 4.0 eV respectively. Find the value of Planck's constant. Solution
We known that
Solution
(i) Binding energy =
13.6
For radiation ofwavelengths
and X2 having energies 5, and
£*2 respectively, we get 0.85 eV =
he
13.6
«,=4
J Ej —£2 ~ he \X]
^„i=0.85eV
13.6 + 10.2 =
c(X2~X,)
^
(4.0I.8)xl.6xia"'^x700xI0^^''>c800xl0~'° (3xl0^)xl00xI0'®
13.6
"2'
l_ X2
. , ^ (£;£*2)^1^2
'Suppose the tfah'sitioh'takes'pldce to' stateTij'for which excitation energyis 10.2eV. In this casegroundstateenergy+ excitation energy= £"«2
he
^.=17 "'"'^2= j
=6.57 X103^ Js.
Solving weget «2 ~ 2 and wehave
^„2=3.4eV Energy ofphoton emitted=• = 3.40.85 = 2.55 eV
Now the wavelength emitted is
# Illustrative Example 1.14
Theionization energyofa hydrogen likeBohratom is4 rydberg. (i)Whatisthewavelength ofradiation emittedwhentheelectron
jumps from firstexcited statetothe ground state ? (ii) What is the radius offirst orbit for this atom ? Given that Bohr radius of
hydrogen atom = 5 x lO"" mand 1rydberg = 2.2x 10~^^ J.
. _ he Solution
1 =>
12431 ,
X=4875A
(i) We know that ionization energy of an hydrogen likeatom is given as
^ " .AtomiG;Physicsi
16
E,_^ =RchZ^
The average.kinetic energy is thus very small comparedto the energybetween the ground state and the'next higher energy state (13.63.4 =10.2 eV).Any atomsin excitedstateemit light and eventually fall to the ground state. Once in the ground
j 12
«)2
=>. ' E,^^ =RchZ'
state, collisions with other atoms can transfer energy of 0.04 eV
Here it is given thatE^_^^ = 4 rydberg = 4 Rch joulethus,we on the average. A small fraction ofatoms can have much more have
Z^=4 =>
Z=2
Thus the wavelength emitted when e~makes a transition from w = 2 to « = 1 is
J
a discharge tube.
I
12
energy(inaccordance witii the distribution ofmolecular speeds), but evenkinetic energy that is 10 tim^ the averageis not nearly enough to excite atoms above the ground state. Thus, at room temperature,nearly all atomsare in the ground state.Atoms can be excited to upper states at very high temperatures or by passing current of high energy electrons through the gas, as in
2^
# Illustrative Example 1.16
^A =10967800x4x 4 • 4
The emission spectrum of hydrogen atoms has two has two
lines ofBalmer series with wavelength 4102 A and4861 A. To
=32903400 A
what series does a spectral line belong if its wave nmnber is equal to the difference of wave numbers of above two lines ?
X.=304A
What isthewavelength ofthisline?[Takei;= 1.097 x lO'm"'] (ii) Radius of
orbit of a hydrogenic atom having atomic
number Z is given as
Solution
r„ =0.529xA
^"
According to given problem
Here Z=2 and n ~ 1 thus radius of first orbit is
^A] =EZ'^
r=0.529x ^A =>
r = 0.2645A
^A2 =RZ^
1
1
«1
«2
1
1
...(1.44)
# Illustrative Example 1.15
Estimate the average kinetic energy of hydrogen atoms (or molecules) at room temperature and use the result to explain whynearly all atomsare in the ground state at room temperature
•^A2 =RZ^
and
and hence emit no light.
2^ in'jf
1
1
1
Xi]
X2
1
RZ^
(ni'f 4
Solution
The wave numbers of above radiations are
Accordingto kinetic theorythe average kinetic energyofatoms or molecules in a gas is given by,
k=\kt
_
1
, 
V, = ^
• Thus
V,

Vi =
1
and V2 = 7— 1.
1
 Z="I X1.38 X1023x300 ' =>
Z =6.2 X102'J.
V[ —V2 =i?Z^
or, in electron volt it is given as 6.2x10 K
—
4
Let the wave number ofthe third emitted line will be
21
1.6x10"^^
(nz'f
=0.04eV. V
=
V, — Vn
...(1.45)
[Atomic Ph^ics ^ 17
AE==(AE)^(AE)^ ...(1.46)
.(V)^ "2i
_5 AE=13.6x^[2/Z,2j 36
Fromequation(l.44),wehave 1
5.667 =I3.6xA[2 2 36 L
II'.
4
„2
(Given thatA5= 5.667 eV) ...(1.47)
=>
«
• .•
4
Applying the law ofconservation ofmomentum, we have, ifP isthe momentum imparted totarget
4 (I.097xl0^)(l)2(4102xl0'®) and
Solving for «2, we get «2 = 6. Similarly from equation(].45), wehave
=4.
i
mgV =2PmgV^
Here
= 2Z^ and
From so, the third line belongs to Brackett series becausethe transition is from level 6 to level 4.
•j =(1.097x10^X1)'
1.
1
L(4)' (6)'J =>
2Z^v =P2Z,v,
and .
Thewavelength isgiven by equation(1.46) so
= IZg asthey contain equal number of
•neutrons and protons. Hence
2ZgV = 2P2ZgVi
=> 2Z^(v+Vi)P •
•
...(1.48)
and 2Zg(v +Vi)2P
.
...(1.49)
Dividing equation(l .48) and(1.49), weget
^=2.62xl0:6j^
If=2 or Z^=2Z,
# Illustrative Example 1.17
...(1.50)
Substituting the value ofZ^ in equation(1.47), we get Two hydrogenlime atoms ^ and B are ofdifferent riiasses, and each atom contains equal number of protons and neutrons.
4VV =3 or Z^=\
...(1.51)
The energy difference between the radiation corresponding to Fromequation(l.50), first Balmerlines emittedby^ and B is 5.667 eV. When the
^.=2Z^=2 atoms Aand B, moving with thesame velocity, strikes a heavy target they rebound back with the same velocity. Inthis process Hence atoms Aand Bare the atom Bimparts twice the momentum tothe target than that respectively.
...(1.52)
(deuterium) and 2He'^ (helium)
A imparts. Identify the atoms ^ and 5.
# Illustrative Example 1,18 Solution
A hydrogehlike atom' ofatomic number Zis in an excited state Theexcitation energies corresponding tothe first Balmer line ofquantum number 2n. Itcan emit amaximum energy photon of (« = 3 to « = 2) emitted by. "i is given by j
£„.^„=(13.6eV)Z^
# Illustrative Example 1.19
A single electron orbits a stationary nucleus ofcharge + Ze, where Z is a constant and e is the magnitude of electronic
1_
...(1.54)
'2^"1
charge. Itrequires 47.2 eV to excite the electron from the second Bohr orbit to third Bohr orbit. Find
Ifw, =2n,the energyofthe emitted photon will be maximum for
(1)
transition W2 = 2 n to«,  1. Thus
(if) The energy required toexcite the electron from the third to 1
E=il3.6eV)Z'
The value of Z
the fourth Bohr orbit.
1
(iii) The wavelength ofelectromagnetic radiation required to
l")
remove the electron from first Bohr orbit to infinity.
^„..=(13.6eV)Z2 Given that £
llfoA
(iv) The kinetic energy, potential energy and the angular
1
momentum to the electron in first Bohr orbit.
An'
(v) Theradius ofthefirst Bohr orbit.
=204eV, weuse 1
204eV=(13.6eV)Z'
...(1.55)
1
An
(The horizontal energyofhydrogen atom =13.6 eV, Bohr radius =5.3 X10""m, velocityoflight=3 XlO^m/s, Planck's constant = 6.6xlO^J.s). •
'
for thetransition ^2 = 2 n to«j = w, we have ✓
1
An^
Solution
1
n^J
Theenergy required to excite the electron from to W2 orbit revolving round thenucleus with charge + Zeisgiven by
Given that£2n_».„=40.8 eV, weuse
40.8eV=(13.6 eV) Z^ f\An^
...(1.56)
=13.6Z2
n
J
eV
"I
«2
Dividing(1.55)by (1.56),we get
(i) According to the given problem, 47.2 eV energy isrequired
1
204
' An^
40.8
r
1
An^
n^
toexcite theelectron from «, = 2 to «2 = 3 orbit. Hence . 472 = z2x 13.6
J 2?. 3.2
\'An' 5 =
2
14
Which gives 4«^ = 16 or« = 2.Using this value of«in either
(1.55) or (1.56) we getZ^ = 16 or Z= 4.
^
47.2x36 _ 25
13.6x5
Z=5.
(ii) Theenergy required toexcite theelectron from k, = 3 to«2 Theground state energy oftheatom corresponds to state « = 1. = 4 orbit is given by Putting n =1 and Z = 4 in equation(1.53), the ground state energy of the atom is
J
^4.£3 = 25X13.6 X
^2
a2
(A)^ £ =_(13.6eV)x •^=217.6eV
, . '
(1)'
Since n = 2, the given excitedatom has
25x13.6x7 = ^ =16,53 eV
= 4.
Itwillemita photon ofminimum energy for a transition /72 = 4 to
(iii) Theenergy required to remove theelectron from the first
«, = 3.Using these values in(1.54), wehave
Bohr orbit to infinity (co) is given by
£».=(13.6eV)(4)^(^^ = 10.58 eV.
£^£, = 13.6xZ
1
L12 £^£,=13.6x25eV
1
cc2.
Momic'.Physics
^
'
Inorder to calculate the wavelength ofradiation, we use
different photons energies. Some ofthe emitted photons have energy2.7 eV, somehaveenergymoreand somehave lessthan
12431
^ 13.6x25 A
2.7eV
(i) Find the principal quantum number ofthe initiallyexcited
X=36.56lA
level 5.
(iv) Weknow for a given orbit for an electron
Total energy =Kinetic energy = ~ Potential energy For
, 19
(ii) Find theionization energy for thegasatoms. (iii) Find the maximum and the minimum energies ofthe emitted photons.
« 1 Solution
Total energy in
orbit is
E=
(i) Figure1.14 shows the energy level Aand Bofhydrogen
13.6Z' eV
like atom. When light ofphoton energy 2.7eV isabsorbed, the electrons go to excited state. Now the atom emits six different
HereZ=5and«= l,thus =I3.6x25eV
=>
.Ei=340eV
Thus kinetic energyin first orbit is K,=+340eV and potential energy in first orbit is
photons such that sixdifferent transitions arepossible. This is only possible when the excited state corresponds to quantum number four. So the principal quantum number of state B must lie between 1 and 4. So either = 2 or = 3. Giventhat
the atoms ofthe gas make transition tohigher energy levels by absorbing monochromatic light ofphoton 2.7 eV. If =3,there will beno subsequent radiations with energy less than2.7 eV. Butradiations with energy less than2.7eVarepossible. Thisis possible when n^l
'
•n = 4 •« = 3
4
Angular momentum ofelectron is mv, r, = 
« = 2
h •
6.63x1034 ' '
2x3.14
'
A
^ Figure 1.14
mv,I '1 r, = 1.055X1034
(v) Radius offirstBohrorbitis givenas
LetZbethecharge onthenucleus ofhydrogen likeatom, then
r, =0.529 x.^A =P
rj =0.529 xiA
=>
ri =0.1058 A
# Illustrative Example 1.20
«=1
E=~Rch —r Now for
n =2
E,— and for
Rch
«=4 RchZ'
A gas ofidentical hydrogen like atoms has some atoms in the
"^4 = 
16
lowest (ground) energy level Aandsome atoms in a particular Thus energy released when electron makes a transition fiom upper (excited) energy level B and there areno atoms in any « =4 ton = 2 is other energy level. The atoms of the gas make transition to higher energy level by absorbing monochromatic light ofphoton RchZ' F F \ 2.7 eV. Subsequently, the atoms emit radiation of only six
l20' Practice Exercise 1.2
E,E^^RchZ^ i_J_ 4 16
=>
(i)
Find theratio ofminimum to maximum wavelength of
radiation emitted byelectron in ground stateofBohr's hydrogen
:3 E^E^ ='^RchZ^
atom.
[3/4]
Further
2.7
RchZ^
(As £4^2 =2.7 eVGiven)
(ii) Determine the wavelength of light emitted when a hydrogen atom makes a transition from « = 6 to « = 2 energy level according to the Bohr model.
^
MZ2 = 14.4eV
[4113.2 A]
(ii) Thus ionizationenergyof atom is •
'•
(iii)
.Ei^^ =MZ^=14.4eV
r. • ,
A double ionized Lithium atom is hydrogenlike with
atomic number 3:
(a) (iii) When electron makes atransition from «=4 then maximum energy is corresponding totransition from « = 4 to « = 1which
Findthe wavelength ofthe radiation required to excite
the electron in Li"^ from the first to the third Bohr orbit.
(Ionization energy ofhydrogen atom equals 13.6eV).
(b)
can be given as
Flow many spectral lines are observed in the emission
spectrum of the above excited system.
i.__L l2
AE_,^Rch'Z^
4?
[114.26 A, 3]
(iv)
'V6
'
'
An energy of68.0 eVisrequired toexcite a hydrogen like
atom from its second Bohr orbit to the third. The nuclear charge is Ze. Find the value ofZ, the kinetic energy ofthe electron in
the first Bohr orbit and the wavelength of the electromagnetic
If
radiation required toeject theelectron from thefirst Bohr orbit to infinity.
max
=tI 14.4 16
[Z = 6, 489.6 eV, 25.39 A.]
AE^ = 13.5eV
(v) The wavelength of the first line of Lyman series for hydrogen is identical tothat ofthe second line ofBalmer series
Similarly the minimum energyis corresponding to transition from « = 4 to w= 3 which can be given as AE_
I A£43 =Rch Z^
AE^^ = RchZ'
9
1
1
16
A^min=i^^l4.4eV
for some hydrogenlike ion X. Calculate energies of the four levels of AT. Also find its ionization potential (Given : Ground state binding energy of hydrogen atom.13.6 eV).. 54.4 eV,  13.6 eV,  6.04 eV,  3.4 eV, 54.4 eV]
(vi) In hydrogen likeatom(atomic numberZ) is in a higher excited stateofquantum numbern. This excited atom can make a transition to first excited state by emitting photons of energies 10.20eVand 17.00eVrespectively. Alternativelythe atomfrom the same excited state can make a transition to the second excited
A£min=0.7eV
; Web Reference at www.Dhysicsgalaxv.com
j Age Group High School Physics  Age 1719 Years
•^iSectionMODERN Topic  Atomic Strticture .. Module Number19 to 36
state by successively emitting two photons of energies 4.25 eV and 5.95 eVrespectively. Determinevaluesof« andZ, (ionization energy ofhydrogen atom = 13.6 eV). [« = 6. 2= 3]
(vii) A gas ofhydrogen like ions is prepared in such a way that ions are only in the ground state and the first excited state.
A monochromatic lightof wavelength 1218 A'is absorbed by the ions. The ions are lifted to higher excited states and emit radiation ofsix wavelength, some higher and some lower than
JAtomic Physics21
theincidentwavelength.Findtheprincipal quantum number of Now the relative picture of atom will be same what we've all the excited state. Identify he nuclear charge on the ions, considered earlier as shown in figure1.16 but electron mass is Calculate the values ofthe maximum and minimum wavelengths, replaced by its reduced mass.

[Z= 2, 4708.71 A, 243.74 A]
, •
rest
1.8 Effect ofMass ofNucleus on Bohr Model
+ Ze
Upto this level we've discussed about the structure of a
hydrqgenic atom by considering nucleus at rest and electron
Figure 1.16
revolving around the nucleus. In fact we know that as no external
force is actingon a nucleus electron ,system, hencethe centre Now we can use all those relations which we've derived for of mass of the nucleus electron system must remains at rest. Bohr model just byreplacing by Such as the radius of Theoretically mass ofelectron isnegligible orsmall compared electron in orbit ofBohr's model is given as
to that of nucleu? and due to this we assume that centre of
mass of the atom is almost situated at nucleus that's why in Bohr atomic model it was assumed that in an atom, nucleus
remains at rest and electron revolves around it.But practically
..,(1.57)
r.=
4K^KZe^m.
But if we consider the motion of nucleus into account, the
the situation is a bit different. Actually centreofmassofnucleus. radius ofn* orbit will be given as
electron system is close to nucleus as it is heavy and to keep
centre of mass at rest, both electron and nucleus revolve around
their centre of mass like a double star system as shown in figure1.15. If r is the distance of electron from nucleus, the
distances ofnucleus and electron from centre ofmass r, and can be given as
+mg) ...(1.58)
ATz^KZe^mgm^
^ mj
r' = y
2
=
...(1.59)
Similarly ifwefind the speed ofelectron in
friMr
and
X
Bohrorbit, it can
be given by equation1.14 as
Wx/+W„
InKZe' + Ze
V
=
nh
Oe
...(1.60)
Here wecansee that in theabove expression ofspeed noterm
Nucleus
ofm^ (mass ofelectron) ispresent, hence itdoes not depend on electron mass, nochange will bethese in speed ofrevolution if we consider the motion ofnucleus into account. + Ze,^2
F,
^HF,
Similarly we know the expression of energyof electron in nth orbit of Bohr model is given as
Figure 1.15
E=
rt'h^
...(1.61)
Taking the motion of nucleus into account the expression of Now we can see that in the atom, nucleus and electron revolve
around their centre ofmass in concentric circles ofradii rj and r2to keep centre of mass at rest. In abovesystemwe can analyze
the motion of electron with respect to nucleus by assuming nucleus to be at rest and the mass of electron replaced by its
energy become E'=
nW
reduced mass u , given as
...(1.62)
mxrm.
nji^+mg
E' =E
...(1.63)
Atomic
Thuswecansaythat the energyof electron willbeslightlyless
10967800 X
compared to what we've derived earlier. But for numerical
= 5483900 m'.
calculations this small change can be neglected unless in a
given problem it is asked to consider the effect ofmotion of
X=
nucleus.
3x5483900
m
X=2430A
# Illustrative Example 1.21
# Illustrative Example 1.22
Calculate the separation between the particles of a system in
the ground state, the corresponding binding energy and wavelength of first line in Lyman series of such a system is positronium consisting of an electron and positron revolving round their common centre.
A p meson (charge  e, mass = 207 m, where misthemass ofelectron) can be captured bya proton to form a hydrogen like 'mesic' atom. Calculate the radius of the first Bohr orbit, the
bindingenergyand thewavelength ofthe firstlinein the Lyman series for such an atom. The mass of the proton is 1836 times
Solution
the mass ofthe electron. The radius of first Bohr orbit and the
The reduced massofthe s>^tem (electron and positron) isgiven
binding energyofhydrogen are 0.529 Aand 13.6 eV respectively. Takeii = 109678 cm'.
by Solution
m.m
2
w + m
i
The reduced mass ofthe system is given by
Where m = mass ofelectron or positron.
(207m)(1836m) The radius offirst Bohr's orbit is given by
^ (207m +1836m) The radius offirst orbit is given by .
r, =
4n^Ke^(m/2)
Tj = 2 Xradius offirstBohr'sorbit ofhydrogen
47c^/:e^(lB6m)
r. =2x0.529 A r, =
r, =1.058A
1
Energy of first Bohr's orbit is given as
=
2K^K^z^e\m/2) ^1 =
0.529
...(1.64)
r, =0.002844A"
From Bohr's theory the ground state energy for hydrogen like atom with Z= 1 is given by
2n^K^e^p
E^=jx13.6qV
=>
Xradius of first Bohr orbit of hydrogen atom
• h'
=  — Xground state energy ofhydrogen
=>
loo
2tr2_4, 2ji^.fi:^e''(186 m)
^1 =
h'
=6.8 eV (Binding energy)
£,=186 X13.6 eV The wavelength of Lyman series is given by £'i=2530eV
j X
X=
= 7?.,
I
1
1^
Hence the binding energy is 2530 eV. The wavelength ofthe Lyman lines are given by
3R^ 2n^K^e'^li
1== p ±JR m
1p «=2,3,4,...
...(1.65)
sAtomlc Phi^ics'
Where
23
= Rydberg constant for mesic atom.
(ii) • The radius of first Bohr orbit is given as h'
Forfirstline i =/fJL_ 1
r, =
'
An^Ke^m
Fr6mequation(1.67), wehave ' %=
...(1.66)
3R..
2A96n'^Ke^m
n^=62A
Now, =>
n =25
=186i?
...
(iii) Rydberg constant for hydrogenic atoms is
.. =186x 10967800m'
Substituting the value ofR^ in equation(1.66), we get 3x186x109678
=>
'
m
ch
=>
An^Ke^m
m
R —
z
ch^ Now when electron is replaced by pmeson, rydberg constant will change as
A,=653.6A. R —
# Illustrative Example 1.23
27i^/:^e\208 m) r—
ch^
R'=208 R=208 X10967800 m^'
A particle ofcharge equal to that ofan electron,  e, and mass 208 times the mass of electron (called a jimeson) moves in a circular orbit around a nucleus ofcharge + 3e. (Take the mass of the nucleus to be infinite). Assuming that Bohr model of the atom is applicable to this system. (i) Derive an expression for the radius of the
Now the wavelength Xofthe radiation emitted when pmeson
makesa transition fromW2 = 3 to « = 1 is .
,
1 • 2
2
«2.
."l
Bohr orbit.
' 1 .1^ 3^
(ii) Find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom.
^=8.R'
(iii) Find the wavelength of the radiationemitted when the pmeson jumps from the third orbit to the first orbit. (Rydberg's constant.^ 10967800 m.) •,
A.=
8R' 1
8x208x10967800
Solution
m
1
^=0.548 A
(i) We have the radius of
orbit of a hydrogen atom as # Illustrative Example 1.24
An^Kie^m
A pimeson hydrogen atom in a bound state pf negatively
Ifelectron is replaced by a heavy particle of mass 20 times that of electron then the radius is given as 2j.2
.rfh r_ =
"
charged pion (denoted byji",
=273 mj and aproton. Estimate
the number of revolutions a 7Cmeson make (averagely) in the ground state of the atom before it decay (mean life ofa 71meson
= 10'® sec), (mass ofproton = 1.67 x,10"^'kg).
AK^K{3)e\20Sm), Solution
r„ =
n'h' 2A96n^Kze^m
...(1.67)
The frequency ofrevolution ofan electron.in hydrogen atom in first orbit is given by
Here we have not used reduced mass because it is given that mass ofnucleus is assumed to be infinite and here we take z=3.
fx
AK^K^e'^m h'
Atomic Physics 1
24
Here in case of a t: meson the mass ofelectron is replaced by the
If effect of mass of nucleus is considered the new value of
reduced mass as
Rydberg constant can be given as
Wp(273 m) mp+273m  ,
2n^K^e^\i ch^
Thus the frequency ofrevolution of%meson will become
RM R
47C^i:e''m^(273 m) h\mp+n3m)
=
m + M
Percentage difference inthevalues ofR"and R^.is given as A/?
4x(3.14)^x9x10^ (1.6x10"'^)'^ x(1.67xl0^')(273x9.1xl0~^')
i?j,i;xlOO
R ~
R..
/, • (6.63x10"'^) 34^3
•^ =^x 100=^0.055% M
(1.67x10"^''+273x9.1x10"^')
=>
. K
y; = 1.77 XID'S sec"'
Thusin the lifetwo'ofn meson 10"^ sec, number ofrevolutions
# Illustrative Example 1.26
made by it is given as Calculate the difference between the ionization potentials of atomic hydrogen and atomic deuterium.
iV=/,xA/ 1.77 X ID'S X 108
=>
Solution
N= 1.77 X 10'® revolutions. '' J
'
The ionization energy for a hydrogenic atom can be given as
illustrative Example 1.25
energy required toexcite electron from Wj = I to «2 ~ given as'
Taking into account the motion of the nucleus of a hydrogen atom, find the expressions for the electron's binding energy in the ground state and for the Rydberg constant. How much (in percent) do the binding energy and the Rydberg constant, obtained without taking into account the motion ofthe nucleus, differ from the more accurate corresponding value of these
1
1
E = Rch 'I
E = Rch
quantities ?
•
1
1
1^
CX.2
E = Rch Joule
*
If motion of nucleus is considered then the value of Rydberg
Solution
constant can be given as
Ifmass ofnucleus is considered (not infinity) then the reduced mass of nucleus electron system can be taken as
R =
mM m + M
Here m is mass ofelectron and Mis that ofnucleus. The binding energy in ground state of hydrogen atom can now be given as
2K'K'e' ch^
Thus the ionization energy of hydrogen and deuterium atoms can be given as 2n'Kh'
£:=13.6x 41 eV
and
2n'K'e'
m
_
=>
^
mMi
.^m +Mf^
E = +
=>
mMi
m + M^
13.6M
(Here M^= 1840 m)
mMjy
\^m +M^ (HereMo=3680m)
The difference of E^^ & Ef^ is
E= ——rr eV m+M
We've hydrogen atom Rydberg constant is given as j . *
.2t^2
4.
2n^K^e^m .R =
ch'
h'
mM 1
mMi
rn + Mj^
m + Mff
Physics
. h'
. =>
3680
1840
3681
1841
on it in the rotating frame ofreference. There may be some
situation in which itis given that electron ofan atom is revolving underthe influence ofa new potential energy field given as
=5.88 X. 10^ J
U=f{r) [Non coulombian field] and using Bohr model, we are
=6.68xl0^^eV
required to develope theproperties ofelectron inthisnewatom.
# Illustrative Example 1.27
For this first we'll develope thefirst postulate equation for this new atom. Aselectron isorbiting inanew potential energy field
A muon isan unstable elementary particle whose mass is 207
centre of orbit given as
givenas U=f{r), it will experience an inwardforce towardthe
and whose charge iseither +eore. Anegative muon (p") can be captured by a nucleus to form a muonic atom. Suppose a
F=
imr)}
dU dr
...(1.70)
proton captures a negative muon (p"), find the radius of first
Bohr orbit and theionization energy ofthe atom. (Take mass of proton = 1836 where ismass ofelectron). Solution
Reduced mass of a system of two particles is given by F =
_ mM _ (207 m^){\ 836 m^) _ m+M " (207m,)+ (1836m^)
\du\
\dr\!
Radius corresponding to thereduced mass p is given by Figure 1.17
m.
•.=
ao =2.85x lOi^m
(^186m,
Thus for a stable
Where Oq istheradius offirst Bohr orbit ofordinary hydrogen
orbit if electron ismoving with speed
in
the orbit ofradius r , we have .
atom. mv.
...(1.71)
fiio =5.29x lO"m.
This equation(1.71) will bethenew equation for first postulate and from Bohr model the second postulate is based on quantization ofangular momentum ofelectron, itsequation will
Ionization energy is given by =186£,
F' —
=>
remain same as
Fi=2.53xl0^eV.
nh
Where £•, istheionization .energy ofordinary hydrogen atom.
1.9 Use ofBohr Modelto DefineHypothetical Atomic Energy Levels
2k
...(1.72)
Nowusing equation(I.71) and (1.72), we can derive all the'
properties for electron motion like, radius of orbit, velocity ofelectron in orbit, angular velocity, fiequency, time period, current, magnetic induction, magnetic moment and the total energy of energy levels for this hypothetical atom in the same
We'vealreadydiscussed that forhydrogenic atoms, Bohrmodel waywe'vederived thesefor properties for a general hydrogenic gives first and second postulates as •• • .i
atom. Lets discuss few examples to understand these concepts
..KZe^ . mvl
in a better way. ...(1.68)
# Illustrative Example 1.28 and
nh mv
r
. " "
~
T—
2%
...(1.69)
Here equation(1.68) is given by balancing .the inward coulombian force on electron bythe outward centrifugal force
Suppose the potential energy between electron and proton at a
distance r is given by  ke^l3j^. Use Bohr's theory to obtain energy levels of such a hypothetical atom.
Atorfiic'PH^iCs'
1;;^ .^ # Illustrative Example 1.29
Solution
According to.given situation for an electron revolving in an «" orbitthe potential energy is given as ...(1.73)
U= .
dU F=
dr
ke' .
...(1.74)
Solution
In the given situation the centripetal force on electron in orbit is given by
' n
orbit electronrevolves at speed
mvl
Bohr's orbit and its energy levels.
3r
The centripetal force on electron due tothis force isgiven as
If in
Suppose potential energy between electron and proton at separation r is given by U= klnr, where isa constant. For such a hypothetical hydrogen atom, calculate the radius of
then we have
ke^
If in
orbit speed of electron is v„ thenwehave mv^
mvl =
ke'
.(1.75)
From Bohr's second postulate, we have nh
According toBohr's qua.ntizatipn postulate, we have nh
(1.76)
 2%
2%
From equation(1.75) and(1.76), wehave 
_
^
'
nh
...(1.80)
...(1.81)
" 2tz4^
Energy ofelectron inn^level is
( nh Y ke^^
E=kE„^PE„
E„= ^mvl klnr
Aiz^ke^m r„ 
' ^
Solvingequation(1.79) and (1.80), weget
lTimr„
and
...(1.79)
mvl  k •
,.(1.77)
21.2 n'h
E^=k\nr 3;.3
n'h and
,.(1.78)
ZTt^km^e^
Now energyin
orbit is
„
I
2
ke
k
'e=2
,,
nh
1ln
C • 2t.2 ^ n h
\^Anmk j
1.10 Atomic Collisions
2
1 ke' , 2 (\ E= 7
6
^
E= r k\n j= 2 •271
2 7.2
ri, h ,
\ A% ke m
n3
In previous sections ofthe chapter we've discussed thatthere are two ways to excite an electron in an atom. One way, of supplying energytoanelectron isby electromagnetic photons, which we've already discussed. We've also discussed that an electron absorbs a photon only when the photon energy is
E
=
equal tothe difference in energies ofthe two energy levels of atom otherwisethe photon will not be absorbed.
iAtpfSe;'ph)«icsi
.271
The another waybywhich energy can be supplied to an elecfron ofanylattice in the colliding body. The energy lost can only be isbycollisions. To understand theenergy supply bycollisions we consider an example of a head on collision of a moving neutron with aStationaryhydrogen atom asshown infigurel. 18.
absorbed by the atom involved in the collision and may get excited orionized bythis energy loss which take place ininelastic collision.
Here for mathematical analysis we can assume the masses of neutron and Hatom as same. neutron
Hatom
H neutron
rest
• Hatom neutron & excited Hatom
(Stationary) Figure 1.18
Figure 1.20
If in thiscase when perfectly electric collision takes place, we In ourcase ofcollision ofa neutron with a Hatom, theenergy know for equal masses here, neutron will come to rest and
hydrogen atom will move with thesame speed andkinetic energy
loss is
m\P (half ofinitial energy ofneutron). This loss in
with which the neutron was moving initially as shown in figurel.19 , .
energy can .be absorbed by Hatom only. We know that the minimumenergyrequiredtoexcitean Hatomis 10.2eV for« = 1
to n = 2. Thushydrogen atom canabsorb onlywhen thisenergy lossis equal to 10.2 eV. Ifthis energy lossin perfectly inelastic collision is morethen 10.2eV then Hatom mayabsorb 10.2eV energyforits excitation and restof the energywillremainin the collidingparticles(w & Hatom) as theirkineticenergyand the
H
collision will not be perfectly inelastic in this case. • neutron
at rest
moving. Hatom
For examplesaythe comingneutron which is going to collide headona stationaryHatom hasinitial kinetic energy 24.5 eV. In this case if perfectly inelastic collision takes place, then the
Figure 1.19
Nowif weconsider the collision to beperfectly inelastic, then aftercollision both neutron andHatom willmove together with speed Vj which can be given by law of conservation of
maximum energy loss can be given as
A£.»=y£,= i(24.5)eV
momentum as
=^.
mvQ = 2 mVj •I
V
...(1.82)
v. = 
In this case the loss of energy can be given as the difference'in initial and final kinetic energies of the neutron and Hatom, given as 
. •
= 1225 eV
From this energy Hatom can absorb either 10.2 eV or 12.09 eV
forits excitation fromk = 1to « = 2 or from « = 1to «= 3 energy level.
Thusin thiscasetheremaybethreepossibilities for,this collision.
u.
These are: AB^E.E
'/
© It maybepossible thatthe collision isperfectly elastic and no energy is absorbed by Hatom as shown in figurel.21 Before collision •
AE=
moving neutron with
AE=^mv'=^E,
K.E. j mv^ =24.5 eV
Hatom at rest
...(1.83)
Thus halfofthe initial kinetic energy will be lost in the collision.
After collision neutron
Oneimportant pointwhich weshouldunderstand carefully is, in collisions ofelementaryparticles and atoms, no energycan be lost as heat because here we can not consider the deformation
at rest
Figure 1.21
Hatom
i/wv2 =24.5 eV
Atomic^ Physics .i
28'
(ii) It maybepossiblethat Hatomwill absorb10.2eV energy during collision and both neutron and Hatomwill be moving withkinetic energy 24.510.2 = 14.3 eVaftercollision asshown in figure1.22. In this casethe collision willbepartiallyelastic.
1 .2:. Finalkineticenergy£^y= ywvJ+,y/«V2 =12.41 eV. ,
In this case the final speeds of neutron and Hatom can.be and energy. Herebymomentumconservation law, we have
Before collision•
Q
7
"
WV = mv, +WV,
moving neutron
Hatom
'
with
at rest
.
K.E.i,„v2.24.5eV
^ "
After collision
•
V
*•
^
'
V=V,+V,

/i o
^
moving neulron with
fnu~2tnv
Hatom at rest
v = u/2
K.E. j ntv' =24.5 eV
' _
...(1.88)
The energy ofexcitation AE is given by n
After collision
T
Vj •
>
A£=
1
o
1
,•
." .o •
y m{ul2)
from n = 1 to /»= 3
 . Figure 1.23
^

•
.
f • 2
AE =. T4'mu
•
• " 
,...(1.89)
i^fomic= [physics
•29
Theminimum.exGitation energy for a Hatom is for transition = 1to = 2 which corresponds to an energy of 10.2 eV.
Let Q be the energylost in the reaction. Then
65 eV =y mv^ +y {4m)v^+Q From equation(1.89)
^ X(1.0078 X1.66 X "• •'
'^
10.2 X(1.6 Xl0'9)
=>
.
zrc T7
1
2 mu^
mv} ^
65eV=Tr/«v,h——
m'=6.24x lOWs. 65eV=
# Illustrative Example 1.31
A neutron ofkinetic energy 65eVcollides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of 90° with respectto its original direction
=>
,
.
(ii) If the atoms gets deexcited subsequently by emitting
65eV =
+^(65 eV)ig
...(1.93)
Now ionization energy of He atom is
(i) Find the allowed values of the energyof the neutrons and that ofthe atom after the collision.
+Q
/£H^+ = 13.6x4 = 54.4eV Thus energy ofelectron in first orbit ofHe atom
./£He+=54.4eV.
.radiation, find the fiequencies ofthe emitted radiation. •
[Given ; Mass of Heatbm= 4 x (massof neutron) lonization energyofHatom= 13.6eV].
Similarly, the energyof electron in second, third and fourth orbits would be—13.6 eV,6 eVand3.4eV respectively. The energy lost must have been used in excitingthe atom.
Solution
First possibility: Let0= 54.413.6 = 40.8 eV
(i) The situation'is shown in figure1.24.
•mv,^ =65x j
= "I" 40.8
= 8 eV nearly
So,
1 ~mvi2i(8eV)=6.4eV
This will be the energy ofrecoil electron.
Neutron
From equation(l .92),
Atom
=>
Figurel.24
ymw^+ymx 16v2^
Applying the law of conservation of momentum, we have •• ..
. , ww = 4w
cos, 0
(1.90)
WV] =4 wv^sin 0
...(1.91)
=4 X~(4m)v2^ 65 eV+6.4 eV = 4 KE ofatom =>
From these equations, we get
= 17.85 eV
Second possibility: Let2=54.46 = 48.4eV.
m4v^cos0 and
KE of atom = ~
Calculating in the same ways as above, we have energy of
V, =4v^sin0
recoil neutron = 0.28 eV
u^\v^\6 Vjv [sin''0 + cos'^ 0]
energyofatom= 16.32 eV
'
= 16
•proceeding as above, let ' u
+v.
16
...(1.92)
2 = 54.43.4=51 eV
=
\ lA.::;Atomic'Phys:rcsj
.30
Inthiscase, theenergy ofrecoil electron= 2.25 eV(negative).
velocity. The atom subsequently returns to its ground state
This is meaningless.
with emission ofradiation ofwavelength 1.216 x 10"' m. Find the velocity of the scattered electron.
Hence
2=40.8eV,48.4eV. Solution
(ii) Itisobvious from first partthattheatom isexcited tothird state or second state. The frequencies of emission can be calculated as follows:
The energylostbythe electron in exciting the hydrogen atom equals the energycorresponding to 1.216 X10"^ m,
n = 3
h = 2
1
u
6.63xl0"^''x3.0xl0^ 1.216x10
"n=l
7
Total 3 Lines
=>
?.= 16.36 xlO'9j
Figure 1.25
Figure1.25 shows the number of emitted frequencies. These
Now, the initial energy ofelectron is20 eV =32 x 10~'^ J.Hence
are
the kinetic energy ofthe scattered electron is
(i)
A]
=RZ''
1
£:=32x i0'9j_ 16.36 xio'^J
1
=>
=(1.097x l0^)x4x
£= 15.64 X10"'^J
The velocity v ofthe scattered electron is given by
= A] =(1.097x 10^)x4x Ax(3xlo8) 4 V, =9.85x1015 Hz 1
(ii)
f 2E^ v =
1/2
I w^
1
19
1/2
9.11x10'^' v=I.86xl0^ms'.
=(1.097xl0^)x4x A A2
2x15.64x10
# Illustrative Example 1.33
y
According to the classical physics, an electron in periodic
V2=
=(1.097 X10')x4 XI X(3 X10^) V2 = 11.7x1015 Hz
(iiO
^3
J 22
32
motion will emit electromagnetic radiation with the same
frequency as that of its revolution. Compute this value for hydrogen atom in quantum state. Under what conditions does Bohr's quantum theory permit emission of such photons due to transitions between adjoining orbits ? Discuss the result obtained.
'=(1.097xl0')x4x ^
V3 ^2^. =(1.097 X10') X4X^ X(3 X10^ = 1.827 xlO'5 Hz
# Illustrative Example 1.32
An electron ofenergy 20 eV collides with a hydrogen atom in the ground state. As a result of the collision,the atom is excited to a higher energystateand the electronis scatteredwith reduced
Solution
The orbital frequency of y
S5
orbit is given by
electron speed  •'
•
"
orbital circumference
"
I Time^ x—X2m/7 So 2ji „2 g ^2 me
rA
1
v„ =
"
47tGo
,
iAtomic,Physics
31
From thelaws of electromagnetic the'ory, the frequency ofthe The energies radiation emitted bythiselectron will also bev^. According to Bohr's theory, the frequency of the radiation for the adjoining orbits is given by E„\
1
me
h
1
12431
12431 ^2=^eV=40.9eV
and v„ =
and E2 of the twoemitted photons in eKare
Thus total energy
(.71)'
£• = .5,+ ^2= 11+ 40.9= 52.3 eV. me
{2n\)
me
(2/tl)
Let Mbe the principal quantum number of the excited state. Using equation(l .94) we have for the transition from m= mto M = l.
v„ =
A'dd 2rp{n~\f
£:=_{54.4eV)
A comparison of this expression with the classical expression above show that the difference in their predictions will be large
1_
But £•=52.3 eV. Therefore
for small«, i.e., for«=2, \lr?= 1/8 and(2m l)/2n^(M1)^ =3/8,
52.3 eV = 54.4 eVx
so that the frequency given by quantum theory is 3 time that calculated from classical theory. However, for very large n, the quantum expression radius to that obtained from classical theory because for m
J 1^
oo 2m1
2m
2m^(m1)^
2}?d
Which gives, m^ = 25 orm= 5.
1
Consequently, the predictions of Bohr's theory agrees with that of the classicaltheory in the limit of verylarge quantum numbers. This correspondence is called as *^Bohr's Correspondence Principle".
The energy ofthe incident electron = 100 eV (given). The energy
supplied to He"*" ion = 52.3 eV. Therefore, the energy of the electron left after the collision = 10052.3 = 47.7 eV.
r WebReference at www.phvsicsgalaxv.com
# Illustrative Example 1.34
! Age Group  High School Physics j Age 1719 Years A100 eVelectron collides witha stationary helium ion(He"^ in
f SectionMODERN PHYSICS
its ground state and excites to a higher level.After the collision,
!• Topic r Atomic Structure
He"^ ions emits two photons in succession with wavelength
! Module Number  37 to 48
1085 A and304A.Findtheprincipal quantum number ofthe excited state. Also calculate the energy ofthe electron after the collision. Given/i = 6.63 x 10"^''Js. Solution
The energy oftheelectron inthem'^ state ofHe"^ ion ofatprnic number Z is given by
£„=(13.6eV),^
Practice Exercise 1.3
(i) Show that for large values ofprincipal quantum number, the frequency of an electron rotating in adjacent energy levels of hydrogen atom and the radiated frequency for a transition between these levels all approach the same value. (11)
Determine the separation ofthe first line ofthe Balmer
series in a spectrum of ordinary hydrogen and tritium (mass
M
number 3).TakeRydberg's constant/? = 10967800 m"'
For He^ion, Z= 2. Therefore
[2.387 A]
(13.6eV)x(2)' £_ = 
77 54.4 E_=—eV
...(1.94)
(ill) Aphoton ofenergy 5.4852 eV liberates an electron from the Li atom initially at rest. The emitted electron moves at right angles to the direction in which photon moves. Find the speed
Atomic Physics
32
iV^, =6.02x 10^Wr'andm^ =9.1 10"^'kg.
momentum and energy to obtaina relation between the minimum distance s ofthe particle from the nucleus in terms of Z, Z, v and b. Show that for ^ = 0, 5 reduces to the distance of closest
[14.2 m/s, 0 = 88.9°]
approach Tq given by
and the direction in which the Li^"^ ion will move. lonization
potential of Li atom = 5.3918 V.Atomic weight of Li = 6.94 g,
1
(iv) A neutron moving with speed v strikes a hydrogen atom in ground state moving towards it with the same speed. Find the minimum speed ofthe neutrons for which inelastic collision may take place. Take mass of neutrons and that of hydrogen
/•„ =
I
2I 4neo
471:
2ZZ'e^ =0
mv
•/'2
ZZ'e
isl.67xlrtg. [3.13 X 10'^ m/s]
(vii)
(v) A uniform magnetic field B exists in a region. An electron projected perpendicular to the field goes in a circle. Assuming Bohr's quantization rule for angular momentum, calculate (a) the smallest possible radius of the electron (b) the radius of the nth orbit and (c) the maximum possible speed of the electron. [(a)
2neB
nh
heB
\ 2neB '
V2nm^ ^
(vO A particle of mass m, atomic number Z, initial speed v and impact parameter b is scattered by a heavy nucleus of atomic number Z. Use the principle of conservation of angular
A small particle ofmass m moves in such a way that the
potential energy U= j
where 6is a constant and r is
the distance of the particle from the origin (Nucleus). Assuming Bohr model ofquantization ofangular momentum and circular
orbits, show that radius of the >7th allowed orbit is proportional
toVw .
,
'
.
"
Advance IllustrationsVideos at www.phvsicsgalaxv.com Age Group  Advance Illustrations Section  Modern Physics Topic  Atomic and Nuclear Physics Illustrations  54 Indepth IllustrationsVideos
^Atomic Physics 33
Discussion Question Q1 1 Balraer series was observed and analyzed before the other Q1 10 Galaxies tend to be strong emitters ofLymanaphotons series. Can you suggest a reason for such an order ? (from the «=2to«= 1transition in atomic hydrogen). But the intergalactic medium—^the very thin gas between the galaxies— Q1 2 You are examining the spectrum ofaparticular gas that is tends to absorb Lymana photons. What can you infer from excited in a discharge tube. You are viewing the discharge these observations about the temperature in these two through a transparent box that contains the same gas. Under environments ? Explain.
what conditions would you expect to see dark lines in the spectrum?
Q1 11 The total energy ofthe hydrogen atom is negative. What significance does this have ?
Q13 The first excited energy ofa He^ ion is the same as the ground state energy of hydrogen. Is it always true that one of Q112 Findoutthewavelength ofthefirst lineoftheHe+ ion the energies ofany hydrogen like ion will be same as the ground ina spectral series whose frequency width isAv = 3.3 x 10^^ s~''. state energy of a hydrogen atom ?
Q1 4 An atom isinits excited state. Does the probability ofits
Q1 13 Very accurate measurements ofthe wavelengths oflight
emitted bya hydrogen atom indicate thatall wavelengths are coming to ground state depend on whether the radiation is slightly longer than expected from the Bohr theory. How might already present or not ? If yes, does it also depend on the the conservation ofmomentum help explain this ?[Hint: Photons wavelengthof the radiation present ? carry momentum andenergy, both ofwhich must beconserved.] Q15 At room temperature, most of the atoms of atomic hydrogen contain electrons that are in the groundstateor « = 1
Q1 14 Inthe Bohr model for thehydrogen atom, the closer the electron is to be nucleus, thesmaller isthetotal energy ofthe
energy level. A tube is filled with atomic hydrogen. atom. Is thisalsotrue in the quantum mechanical picture of the Electromagnetic radiation with a continuous spectrum of hydrogen atom ? Justifyyour answer. wavelengths, including those in the Lyman, Balmer, andPaschen series, enters one end of this tube and leaves the other end.
Theexisting radiation is found to contain absorption lines. To which one(or more) ofthe series dothe wavelengths ofthese absorption lines correspond ? Assume that once an electron
Q115 The materials (phosphors) that coat the inside of a
fluorescent lamp convert ultraviolet radiation (fiom the mercuryvapor discharge inside the tube) into visible light. Could one alsomake a phosphor thatconverts visible lighttoultraviolet ?
absorbs a photon and jumps to a higher energylevel, it does not absorb yet another photon and Jump to an even higher
Ejq3lain.
energy level. Explain your answer.
Q1 16 Consider the line spectrum emitted from agas discharge
tube such as a neon sign or a sodiumvapor or mercuryvapor Q16 When electromagnetic radiation is passed through, a lamp. Itisfound thatwhen the pressure ofthevapor isincreased, sample ofhydrogen gasat room temperature, absorption lines the spectrum lines spread out over a larger range of areobserved in Lyman series only. Explain. wavelengths, that is, areless monochromatic. Why ?
Q1 7 The difference in the frequencies ofseries limit ofLyman Q1 17 Which wavelengths will be emitted byasample ofatomic series and Balmer series is equal to the frequency ofthe first hydrogen gas (in ground state) ifelectrons ofenergy 12.2 eV line ofthe Lyman series. Explain.
collide with the atoms ofthe gas ?
Q18 When an electron goes from the valence band to the
Q118 As a body is heated to a very high temperature and
conduction band in silicon, its energy is increased by 1.1 eV.
becomes selfluminous, the apparent color of the emitted radiation shifts from red to yellow and finally to blue as the
The average energyexchanged in a thermal collision is ofthe order of A:/" which is only 0.026 eV at room temperature. Howis a thermal collision able to take some of the electrons from the
temperature increases. Why the color shifts ? What other changes in the character of the radiation occur ?
valence band to the conduction band ?
Q19 Doesit takemoreenergyto ionize(free) the electron ofa hydrogen atom that is in an excited statethan onein the ground state ? Explain.
Q119 Explain whythe Bohr theory is applicable onlyto the hydrogen atom and to hydrogenlike atoms, such as singly ionized helium, doublyionized lithium, and other oneelectron systems.
.34
^
^c• „
Q1 20 What are the most significant differences between the Bohr model ofthe hydrogen atom and the Schrodinger analysis ofthat atom ? What are the similarities ?
Q124 The numerical value of ionization energy in eVequals
theionizationpotential in volts. Does the equality hold ifthese quantities are measured in someotherunits ?
Q121 Howmany wavelengths are emitted byatomic hydrogen Q125 When the outermost electron inan atom isin an excited in visible range (380 nm 780 nm) ? In the range 50 nm to .state, the atom ismore easily ionized than when theoutermost 100nm ?
electron is in the ground state. Why ?
Q122 Stars appear to have distinct colors. Some stars look red, someyellow, and othersblue.What is a possibleexplanation
Q126 Elements in the gaseous state emit line spectra with welldefined wavelengths. But hot solid bodies always emit a continuous spectrum, that is, a continuous smear of wavelengths. Can you account for this difference ?
for this?
Q123 What will be the energycorresponding to the first excited state of a hydrogen atom if the potential energy ofthe atom is taken to be 10 eV when the electron is widely separated from
the proton ?Can we still write
=E^/n^ ? ~
iAtomic; Physics
35,
ConceptualMCQsSingle Option Correct 11 According to Bohr's theory of the hydrogen atom, the 18 Inthefollowing figure! .26 the energy levels ofhydrogen total energy ofthehydrogen atom with itselectron revolving in atom havebeen shown alongwith sometransitions marked A, the «th stationary orbit is : B, C,Z)and£.Thetransitionsy4,£andCrespectivelyrepresent: (A) (B) (C) (D)
Proportional to « Proportional to Inversely proportional to « Inversely proportional to
OeV
« = 5
•.544eV
n = 4
12 According to Bohr's theory of the hydrogen atom, the radii ofstationary electron orbits arerelated totheprincipal
0.85 eV
B
« = 2
cc l/«2
(B)
(Q r cc«
cc 1/k
•D
A
quantum number « as :
(A)
E
•C
n = 3
n=\
1.5eV
•  3.4 eV
—13.6 eV
p) r Figure 1.26
13 The wavelengths involved in the spectrum of deuterium (j D) are slightly different from that of hydrogen spectrum,
(A) The first member of Lymanseries,third member ofBalmer
because'
series and second member of Paschen series
(A) (B) (Q p)
.
.
Size ofthe two nuclei are different Nuclear forces are different in the two cases Masses ofthe two nuclei are different Attraction between the electron and the nucleus is different
in the two cases
P) The ionisationpotential ofhydrogen, second member of Balmer series and third member of Paschen series
(Q The series limit ofLyman series, second member ofBalmar series and second member of Paschen series
.
''
P) The series limit of Lyman series, third member of Balmer series and second member of Paschen series
14 The energy of the electron of hydrogen orbiting in a stationary orbit of radius is proportional to :
(A)
.
(Q
19 The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following
P) l/r„
statements is true ?
(D) l/^„
(A) Its kinetic energy increases and its potential and total energies decreases
15 The shortestwavelengthof the spectrumfor transition of an electron to« = 4 energylevelofa hydrogen likeatom(atomic number = 2) is the same as the shortest wavelength of the Balmer series ofhydrogen atom. The value ofZis:
(A) 2
P) 3 . " " •
(Q 4
P) 6
16 Which of the following series in the spectrum of the hydrogen atom lies in the visibleregion of the electromagnetic spectrum ? (A) Paschen series (Q Lyman series
P) Balmer series p) Brackett series
17 The angular momentum of an electron in an orbit is quantized because it is a necessary condition for the compatibilitywith: (A) The wave nature of electron p) Particle nature ofelectron (Q Paulli's exclusion behaviour p) None of these
p) Its kinetic energy decreases, potential energy increases and its total energy remains the same
(Q Its kinetic and total energies decrease and its potential energy increases
P) Its kinetic,potential and total energies decrease 110 In the Bohr's model ofhydrogen atom, the ratio of the kinetic energy to the totalenergy oftheelectron in quantum state is:
(A)l ' (Q 2
.
'
•
• p) +1 p).2 ^
111 According to Bohr's theory of the hydrogen atom, the speed ofthe electron in a stableorbitisrelatedto theprincipal quantum number n as (C is a constant) :
(A) (Q v^=Cxn
p) v„=C/h p) v„ = Cx«2
112 In the Bohr model of a hydrogen atom, the centripetal forceis furnished by the coulombattraction betweenthe proton and the electron. If is the radius ofthe ground state orbit, m
AtornicPhysics
36%
is the mass and e is the charge on the electron and Bq is the vacuum permittivity, the speed ofthe electron is : (A) 0
(B)
(Q
(D)
118 The difference in angular momentum associated with electron in two successive orbits of hydrogen atom is :
(A)f (C)f
^AnSQaQm
119 If radiation ofall wavelengthsfromultraviolet to infrared 113 If elementswith principal quantumnumber n > 4 werenot allowed in nature, the number of possible elements would be : (A) 60 (B) 32 (Q 4 (D) 64 114 ^Bohr's atomic model gained acceptance above all other models because it:
(A) Is based on quantum hypothesis (B) Explained the constitution of atom (Q Assumed continuous radiation of energy by orbiting
is passed through hydrogen gasat room temperature absorption lines will be observed in the
(A) Lymanseries (Q Both (A) and (B)
(B) Balmerseries p) Neither (A) or (B)
120 Which ofthe following force is responsible for aparticle scattering ? (A) Gravitational (B) Nuclear (Q Coulomb p) Magnetic
electrons
121 A Hydrogen atom and Li"*^ ion are both in the second
(D) Explained hydrogen spectrum
excited state. If£^and aretheir respective angular momenta, andE^ andE^^ their respective energies, then:
115 Pauli's exclusion principle states that'no &vo electrons in an atom can have identical yalues for: (A) One of the four quantum nurnbers ^ (B) Two ofthe four quantum numbers (Q Three ofthe fouf quantum numbers (D) All four quantum numbers
(A)
and
P) (Q
= 1^1 and = 1^1 and
p)
and
116 Energy levels A, B and C ofa certain atom correspond to
increasing values ofenergy i.e. E^E^>E2
123 The wavelength ofradiation emitted due to transition of electron from energy levelE to zero is equal to X. The wavelength
ofradiation (X.j) emitted when electron jumpsfrom energy level 3E ^to zero will be:
h
Figure 1.27
(A)
=
+^
(C) X,] + X2 + X3 = 0
(D)
+A.'
117 When white light (violet to red) is passed through hydrogen gas at room temperature, absorption lines will be observed in the
(A) Lyman series (Q Both (A) and (B)
(B) Balmer series (D) Neither (A) or (B)
Figure 1.28
(A) fx
(B) fx
(Q
(D)
!;Atomic_Physics
,
124 A neutron collies headon with a stationary hydrogen atom inground state. Which ofthe following statements is/are correct ?
128 Hydrogen H, deuterium D, singly ionized helium He"^ and doubly ionized lithium Li"^all have one electron around the
(A) Ifkinetic energy ofthe neutron isless than 13.6 eV, collision
nucleus. Consider n= 2and w= 1transition. The wavelengths ofthe emitted radiations areXj, and X^ respectively. Then
must be elastic
approximately:
(B) Ifkinetic energy oftheneutron isless than 13.6 eV, collision maybe inelastic
(Q Inelastic collision maytakeplace on when initial kinetic energy ofneutron is greater than 13.6 eV
(D) Perfectly inelastic collision cannottake place
(A) X^ =1X^=242X^=2>42X^ (B) X^ =X^ =2X^ =2,X^ XQ X, = X,=4X^ = 9X, . P) 4X^ =2X^=2X^ = X^
125 An electron inhydrogen atom after absorbing an energy 129 Which ofthe following curves may represent the energy photon jumps from energy state to n^. Then it returns to ofelectron inhydrogen atom asa function ofprincipal quantum ground state after emitting six different wavelengths inemission
number n:
spectrum.'The raergy ofemitted photons iseither equal to, less
than or graterth^ theabsorbed photons: Then wj,and (A) = =3 (B) 02 =5,«j3
(C) ^2 =4,"1 =2
(D) «2=4,nii
1 26 Mark correct statements:
(A) Bohr'stheoryis applicable to hydrogen alonebecause its nucleus is most light
(B) Binding energy ofelectron (inground state) of isgreater than that of jHin ground state (Q Allthe linesofBalmerseries lie in visible spectrum p) None of these
127 Figure1.29 represents transitions ofelectrons from higher tolower state ofa hydrogen atom. Which transition represents the line ofBalmer series:
(C) 0 E
« = 4'
n = 3
130 Ahydrogen atom inground state absorbs 12.1 eV energy. Theorbital angular momentum ofelectron is increased by:
m= 2
(B) f (9 i
«=I
P) zero
Figure 1.29
(A) 1 (C) 3
(B) 2 P) All 1,2, and 3
131 Magneticmomentdueto the motionofthe electron in energy state of hydrogen atom is proportional to : p) (A) n (C) p) ,,3
Atomic Physics
NumericalMCQs Single Optioris Correct 11 Theenergy required to excite a hydrogen atom from n = 1 18 The different lines in the Lyman series have their to « = 2 energy state 10.2 eV. What is the wavelength of the wavelengths laying between: P) 900 Ato 1200 A radiationemitted bythe atom when it goesback to its ground (A) Zero to infinite
(C) 1000 Ato 1500 A
state ?
(A) 1024 A (Q 1218A
(B) 1122 A P) 1324A
p) SOOAtolOOOA
19 The orbitalelectron of the hydrogenatomjumps fromthe
ground state to a higher energy state and itsorbital velocity is
12 ConsiderBohr'stheoryfor hydrogen atom.The magnitude reduced to one third ofits initial value. Ifthe radius ofthe orbit ofangular momentum, orbit radius andfrequency oftheelectron in the ground state is r, then what is the radius of the new orbit ? in energystatein a hydrogen atom areL, r &/respectively. (A) 2r P) 3r Find out the value of 'x', if the product f r L is directly (Q 4r P) 9r proportional to if: (A) 0 (C) 2
(B) 1 P) 3
.
110 If we assume that penetrating power of any
radiation/particle is inversely proportional to its deBroglie
13 For the first member ofBalmer series ofhydrogen spectrum,
the wavelength is X. What is the wavelength of the second member?
(A) 30^
(B) YgX
(Q 1?^
(D) fx
wavelength of the particle then :
(A) a proton andan aparticleafter getting accelerated through samepotential difference willhave equal penetrating power. P) penetrating power ofaparticlewillbe greater thanthat of proton which havebeen accelerated bysame potential difference. (Q proton's penetrating power will be less than penetrating power of an electron which has beenaccelerated bythe same potential difference.
p) penetrating powers can not be compared as all these are particles having no wavelength or wave nature.
1*4 In a new system of units the fundamental quantities are
planks constant Qi), speedof light (c) and time (7). Then the dimensions of Rydberg'sconstant will be :
(A) (Q
excited states ofhydrogen will be: (A) 1:8 P) 8:27
P) h^c^r^ P) /z"' c P
c® r'
111 According to Bohr's theory the ratio of time taken by electronto completeone revolutionin first excitedand second
(Q 8^:272
P) 4:9
15 In a hypothetical atom, if transition from « = 4 to w= 3 produces visible light then the possible transition to obtain 112 The area of the electron orbit for the ground state of hydrogenatom is^. What will be the area of the electronorbit infrared radiation is: (A) « = 5 to « = 3 (Q « = 3 to w= 1
corresponding to the first excited state ? (A) 4 A P) 8A
P) rt= 4 to « = 2 p) none of these
(Q 16A
16 If first excitation potential of a hydrogen likeatom is Velectron volt, then the ionization energy ofthis atom will be: (A) V electron volt W
AV
17 lonisatiorienergy for hydrogen atom in the ground state is
E. Whatisthe ionisation energy ofLi"^ atomin the.2"'' excited state ?
•.
(A) E
P) 3£ 
hydrogen is
(G) n = 6
electron volt
P) cannot be calculated by given information
(C) 6E
113 Ifthe wavenumber of a spectral line ofBrackett series of 9
times the Rydbergconstant. What is the state
from which the transition has taken place ? (A) « = 4 p) « = 5
P) — electron volt (Q
P)'32A
p) 9:E
.
"•
p) n = l
Ionization potential ofhydrog^ atom is 13.6 V. Hydrogen atom in the ground state is excited by monochromatic radiation ofphotons of energy12.09 eV. The number of spectral lines emitted by the hydrogen atom, according to Bohr's theory,will be:,
(A)Pne;.
'•(^;';i[iree
' / p) Two •
P)^dur
„" •'
LAtomic Physics
3£f5
115 An electronjumps from the first excited stateto the ground 122 In a hydrogen atom following the Bohr's postulates the stage ofhydrogen atom..What will be the percentage change in product of linear momentum and angular momentum is the speed ofelectron ?
(A) 25% (Q 100%
proportional to/T* where '«'is the orbit number. Then 'xMs:
•
(B) 50% (D) 200%
(A) 0 (Q 2
(B) 2 (D) 1
electron from the neutral helium atom. The energy(in eV) required
123 The velocity ofan electron insecond orbit oftenlyionized sodium atom (atomic number Z= 11) is v. Thevelocity ofan
to remove both the electron from a neutral helium atom is:
electron in its fifth orbit will be:
(A) 382 (Q 51.8
(A) V
(B) fv
(C) v
22 (D) fv
116 An energyof 24,6 eV is required to remove one of the
(B) 492 (D) 79.0
117 A neutron beam, in which each neutron has same kinetic
energy, is passed through a sample of hydrogen likegas (but not hydrogen) in ground state and at rest. Due to collision of neutrons with the ions of the gas, ions are excited and then they emit photons. Six spectral lines are obtained in which one
ofthe linesis ofwavelength (6200/51) nm. Which gasis this ? (A) H (Q
(B) He^ (P) Br3
118 In previous questicm whatis theminimum possible value of kinetic energy of the neutrons for this to be possible. The mass of neutron and proton can be assumedto be nearly same. Use
=12400 eVA.
(A) 31.875eV (C) 127.5 eV
(B) 63.75eV (D) 182.5 eV
124 A positronium consists of an electron and a positron revolving about their common centre of mass. Calculate the
separation between the electron and positron in theirfirstexcited state:
(A) 0.529A (Q 2.116A .
(B) 1.058A p) 4.232A
125 A positronium consists of an electron and a positron revolving about their common centre of mass. Calculate the
kinetic energy of the electron in ground state: (A) 1.51 eV (B) 3.4ey (Q 6.8eV (p) 13.6eV
126 A hydrogen atom is in an excited state of principle quantum number n. It emits a photon of wavelength X when
119 In Millikan's oil dropexperiment, a chargedoil dropof mass 3.2 X10"^'* kg is held stationary between two parallel
returns to the ground state. The value ofn is :
plates 6mm apart by applying apotential difference of1200 V (A) ^XRCkRl) between them. How many excess electrons does the oil drop carry? Takeg = 10ms~^: (A) 7 (Q 9
(B) 8 (D) 10
(Q
(B)
XR
(>J?1) XR
(D)
XR\
127 Theratio ofmagnitude ofenergies ofelectron inhydrogen
120 In a hydrogen like atomthe energyrequiredto excitethe
atom in first to second excited states is :
electron from 2"*^ to S"' orbit is 47.2 eV. What is the atomic
(A) 1:4 (Q 9:4
number ofthe atom ?
(A)2
.
(B) 4:9 P) 4:1
....
(B).3
128 Monochromatic rafriatipn ofwavelength Xisincident on a hydrogen sample in ground state. Hydrogen atoms absorba .fractionoflight and subsequentlyemit radiations ofsix difierent 121 Twohydrogenatoms are in excitedstate with electronsin wavelengths. Find the wavelength X: w= 2 state. First one is moving towards left and emits a photon.' (Q 4
of energy
(P) 5

towards right. Second one is moving towards
right with same speed and emits aphoton ofenergy £"2 towards right. Taking recoil of nixcleus.into account during_emission process : . . / . . •
(A).£j>£2 (C) £j£'2'
(A)975 A (C) 2248A
.
(B) 1218A P) 4316A
129 Out ofthe following transitions, the frequency ofemitted photon will be maximurn for.:
(B) ./ • •(A)V«^'5t6n'^.3.. • r;. p) n = 6tOff = 2 (D) information insufficient  : ;(Q.'«^23p«i';.p
Atomic Physics ;
40:
130 Imagine a neutral particle of same mass m as electron revolving around a proton of mass only under newton's gravitational force. Assuming Bohr's quantum condition, the radius ofelectron orbit is given by: (A)
atom equal to: 1
7
(B)
3Z^R
(B)
nm^GMp
16
(Q
2 72 GMpN^h
(C)
The longest wavelength of photon that will be emitted has wavelengthX(given in terms of Rydberg constant^? ofhydrogen
nhGM„
P)
4n^m^
^ 4Km
131 Determine the ratioofperimeters in 2"^ and 3"^ Bohrorbit in He^ atom:
P)
3Z^R
3Z^i?
137 Oneofthe linesin the emission spectrum ofLi2"*" hasthe same wavelength as that of the 2"^^ line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is:
(A) f (Q I
(A) n = 4^/7 = 2 (C)« = 8^n = 4
.(B) ig
(D)f
P)n = 8^rt = 2 P) n=12^« = 6
138 In Bohr's theory the potential energy of an electron at a Kr'
132 The photon radiated from hydrogen corresponding to
position is
2"^^ lineofLyman series isabsorbed bya hydrogen likeatom 'A" in 2"^ excited state. Asa result thehydrogen likeatom 'AT makes
quantized energy of the electron in nth orbit is:
a transition to
orbit. Then,
(A) X= He+, n = 4 (Q ArHe^« = 6
(B) AT Li++,« = 6 (D) A'=Li^,«9
133 An aparticlewith a kinetic energyof2.1 eV makes a head on collision with a hydrogen atom moving*towards it with a kinetic energy of8.4 eV.The collision : (A) must be perfectly elastic (B) may be perfectly inelastic (Q maybe inelastic P) must be perfectly inelastic
(A)
nh f K 2n\m
(q n{^]
(where A!" is a positive constant); then the
,1/2
(B)
P)
2jcl^m J nh (
1/2
2iz{k)
139 Iffirst and second frequencies in transition to
orbital
are relatedbythe relation Vj = kv^, then the first frequency in the transition to second orbital will not be equal to :
(A) V. (il)
P) (1A)V2
134 A hydrogen atom is initially at rest and free to move is in the secondexcitedstate. It comesto ground state by emitting a
(Q V2Vj
P)
photon, then the momentum of hydrogen atom will be approximately: (in kgm/s)
140 The ratio of deBroglie wave length of a photon and an electron ofmass 'w' having the same kinetic energy £ is: (Speed oflight = c)
(A) 12.1x027 (Q 3 X102^
p 6.45x1027 P) 1.5 X027
135 An gas ofHatoms in excited state
absorbs a photon
(A)
2mc
P)
mc
\
E
of some energy and jump in higher energy state n^. Then it returns to ground state after emitting sbcdifferent wavelengths in emission spectrum. The energy of emitted photon is equal,
(Q
2mc
mc
P)
~Y
less or greater thanthe energy ofabsorbed photon thenWj and will be:
(A) Kj =5, «2 = 3 =4,«2~3 (Q
P) Wj=5, «2~2 P) =4, M2~2
136 Imagine anatom made of a nucleus ofcharge (Ze) and a hypothetical particle of same mass but double the charge of the electron. Apply the Bohr atom' model and consider all possible transitions of this hypothetical particle to the ground state.
141 A monochromatic radiation of wavelength Xis incident on a sample containing He"*". As a result the Helium sample starts radiating. A part of this radiation is allowed to pass through a sample of atomic hydrogen gas in ground state. It is noticed that the hydrogen sample has started emitting electrons whose
maximum Kinetic Energy is37.4 eV. (he = 12400 eVA)Then Xis: (A) 275 A P) 243 A (Q 656A P) 386A
jAtomic Physics
142 Ofthefollowing transitions in hydrogen atom, theone which gives an absorption line ofhighest frequency is : (A) n = \ton = 2 (Q « = 2to/7=l
(B)« = 3tort = 8 (D)« = 8to« = 3
146 An orbital electron in the ground state ofhydrogen has an angular momentum L,, and an orbital electron in the first
orbit in the ground state oflithium (double ionised positively) has an angular momentum ZjThen; (A)Zj=Z2 (B)Z, =3Z2 143 An electron ofthe kinetic energy lOeV collides with a (C) ^2 =31, (p) L^ =9L^
hydrogen atom in 1st excited state. Assuming loss of kinetic
energy inthe collision tobe quantized which ofthe following statements is INCORRECT.
(A) Thecollision maybeperfectly inelastic (B) The collision may be inelastic (Q The collisionmay be elastic
147 The ratio ofthe maximum wavelaigth ofthe Lyman series in hydrogen spectrum to the maximum wavelength in the Paschen series is:
(A)
p) The collision must be inelastic
144 Ifin the first orbit ofahydrogen atom the total energyof
3 105
(Q ~
(B)
108
theelectron.is21.76,x lQ'%then itselectric potential energy will be:
'
(A) 43.52 X1019 J (C) 10.88x1019;
" r '
(B) 21.76x1019; p) 13.6x1019;
145 In thefigure sixlines ofemission spectrum are shown. Which ofthem will be absent in theabsorbtion spectrum.
4 1
2
5
X
(Q 4,5,6
Suppose Zj, £2and£3areminimum energies required sothat the atoms H, He% Li"*"*" can achieve their first excited states respectively, then:
(A) £,=£2 =£3 (Q E^£3 p) £,=£2 =£3
149 The radius offirst Bohr orbit ofhydrogen atom is0.53 A. Then the radius of firstBohrorbit of mesonic atom (negative meson hasmass 207times thatofelectron butsame charge) is : (A) 2.85X lOi^m (B) l.ObxlOi^m (Q 0.53xloiOm
3
Figure 1.30
(A) 1,2,3
148 Consider atoms H, He^, Lif^ in tfieif'^buny states.
(P) 7.0xioi2m
AtomicPhysics
42
AdvanceMCQs with One or More Options Correct 11 The ground state and first excited state energies of 17 Which ofthe following transitions inHe"*" ion will give rise hydrogen atom are13.6 eVand3.4 eVrespectively. Ifpotential toa spectral line which has the same wavelength assome spectral line in the hydrogen atom ? energy in ground state is taken to be zero. Then : (A) potential energyin the firstexcited statewould be20.4eV (A) K= 4tow = 2 P)« = 6ton = 2 total energyin the first excitedstate wouldbe 23.8 eV (Q « = 6 to « = 3 p) n = 8 to « = 4 (Q kineticenergy in the first excitedstate wouldbe 3.4 eV P) total energyin the ground state wouldbe 13.6eV 18 In the Bohr model of the hydrogen atom, let R, Vand E represent the radius ofthe orbit, speed ofthe electron andthe 12 An electron is excited from a lower energy state to a higher magnitude of total energy of the electron respectively. Which energy state in a hydrogen atom. Which of the following of the following quantities are proportional to the quantum quantity/quantities decreases/decrease in the excitation ? number «? (A) Potentialenergy (B) Angular speed (A) VR (^) RE (Q Kinetic energy
P) Angular momentum
13 An electronin a hydrogenatom makes a transition
{D)
(Q f
where nj and ^2 ^re principle quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is eight times that in the final
19 In an electron transition inside a hydrogen atom, angular momentum ofelectron may change by
state. The possible values of and «2are • (A)«j4,«2 = 2 P) /7^ = 8,W2 = 2
(A) h
(Q nj=8,«2~l
P) «i=6,«2^2
14 An electron in hydrogen atom first jumps from second
P) 7C ^ h
(Q
2k
(D) A 471:
excited state to first excited state and then from first excited
state to ground state. Let the ratio of wavelength, momentum and energy of photons emitted in these two cases be, a, b and c respectively. Then:
(A)c=i
P) fl= 9/4
(C) 6 = 5/27
P) c=5/27
110 A beam of ultraviolet light of all wavelength passes
through hydrogengas at roomtemperature, in thexdirection. Assume that all photons emitted due to electron transitions insidethe gas emergein the ydirection. Let^ and B denotethe lights emerging from the gas in the xand ydirections respectively:
15 The magnitude of energy, the magnitude of linear
(A) Someof the incident wavelengths will be absentin A P) Only those wavelengths will be present in B which are
momentum and orbital radius ofan electron in a hydrogen atom
absent in A
corresponding to the quantum number n are E, P and r respectively. Thenaccording to Bohr's theoryofhydrogen atom :
(Q B will contain some visible light P) B willcontainsomeinfraredlight
(A) EPr is proportional to —
111 Whenever a hydrogenatom emits a photon in the Balmer
P) P/E'is proportional to « (Q Er is constant for all orbits P) Pr is proportional to n
series:
(A) It may emit another photon in the Balmer series p) It miist emit another photon in the Lyman series (Q The secondphoton, if emitted, will have a wavelength of
16 The wavelengths and frequencies of photons in transitions 1,2 and 3 for hydrogen
about 122 nm
(b) It may emit a second photon, butthewavelength ofthis photon cannot be predicted
likeatom areX.,, Xj, >13, vj, V2 and V3 respectively. Then : (A) V3 = Vj+V2 v,v
P) v,=
(Q ^ ~
V777777777777777777777777Z,
Figure I.OO
V1+V2
^
X.iX,2
112 Which of the following statements about hydrogen spectrum is/are correct ? (A) All the lines ofLyman series lie in ultraviolet region P) All the lines ofBalmer series lie in visible region
(C) All the lines of Paschen series lie in infrared region P) none of these
[Atomic Physfcs
43:
113 Aneutron collides headon with a stationary hydrogen
118 Suppose the potential energy between electron and
atomin groundstate. Which ofthe following statements is/are correct ?
proton at adistance r is given by^^.
(A) ifkinetic energy oftheneutron isless than 13.6 eV, collision
2r^ Using Bohr's theory choose the correct statements :
must be elastic
(B) ifkinetic energyoftheneutron is less than 13.6 eV, collision
(A) Energy in the nth orbit is proportionalto (B) Energyin the nth orbitis proportional to
maybe inelastic
(C) Energy is proportional to rrp {m :mass ofelectron)
(Q inelastic collision maytakeplaceonlywheninitialkinetic p) Energy isproportional to
(m: mass ofelectron)
energy ofneutron is greater than 13.6 eV
(D) perfectly inelasticcollision cannottakeplace
114 If, in hydrogen atom, radius of frequency ofrevolution ofelectron in
119 An electron in an hydrogen atom has total energy of
Bohr orbit is r^, orbit is^ and area
enclosed by w''* orbit is A^, time period ofelectron is
then
which of the following graphs is/are correct ?
Mi. (A)
120 When Z is doubled in an atom, which of the following statements are consistent with Bohr's theory ? (A) Energy of a state is double
(B)
logn log
(Q
3.4 eV. Choose the correct statement (s): (A) The kinetic energy ofthe electron in that orbit is 3.4 eV P) The potential energy ofthe electronin that orbit is 6.8 eV (Q Angular momentum ofthe electron in that orbit is h/n p) Angular momentum ofthe electron for that orbit is 2h/%
IT
P) Radius of an orbit is doubled (C) Velocity ofelectrons in an orbit is doubled p) Radius of an orbit is halved
121 The electron in a hydrogen atom jumps back from an excited state to ground state, by emitting a photon of 16 ^ wavelength , whereis Rydberg's constant. In place
(D)
15 K
logn
ofemittingonephoton, the electron couldcomebackto ground state by
115 Mark correct statement(s): (A) Bohr's theory is applicable to hydrogen alone because its nucleus is very light
(A) Emitting 3photons ofwavelengths
and ^3 such that
15R 16
^2 ^3 (B) Binding energy ofelectron (in ground state) ofjrf is greater (B) Emitting 2 .photons of wavelength A,,and than that ofjH' in ground state
such that
(Q all the lines ofBalmer series lie in visiblespectrum
J_
1
(D) none of these
A,
X2 ~ 16
116 A photon of energy 10.5 eV isallowed to interact with a hydrogen atom in its ground state. Then :
(C) Emitting 2 photons of wavelength X^ and X^ such that
15R
16
(A) the photon is completely absorbed bythe Hatom (B) the photon cannot excite the Hatom and comes out with energy 10.5 eV (Q thephoton transfers 10.2eV energyto Hatomexcitingit to
(D) Emitting 3photons ofwavelength Aj, Aj andA3 such that 16
first excited state
P) none of these
122 The photon radiated from a hydrogen atom corresponding
117 When a hydrogen atom is excited from ground state to first excited state:
(A) Its kinetic energy increases by 10.2 eV (B) Its kinetic energy decreases by 10.2 eV (Q, Its'potential energy increases by 20.4 eV
(D) Itsangular momentum increases by1.05 x lO'^^ Js
to 2"^^ line ofLyman series isabsorbed by a hydrogen like atom 'JT in 2"^ excited state. Asa result thehydrogen likeatom 'X" makes a transition tow'^orbit. Then predict 'X and the quantum number:
(A) He"',«= 4 (Q X=He\n = 6
(B) X=Li^,n = 6 (D) X=Li^,n^9
Atomic Physics ;
44
123 A particular hydrogen like atom has its ground state (A) Ifkinetic energy oftheneutron islessthan20.4eVcollision must be elastic binding energy 122.4 eV. It is in ground state. Then p) Ifkineticenergyoftheneutronislessthan 20.4eVcollision (A) Its atomic number is 3 maybe inelastic (B) An electron of90eV can excite it (Q An electron ofkineticenergynearly91.8eVcan bebrought (Q Inelastic collision may be take place only when initial kinetic energy ofneutron is greater than 20.4 eV to almost rest by this atom P) An electron of kinetic energy 2.6 eV may emerge from the P) Perfectly inelastic collision can not take place atom when electron ofkinetic energy 125 eV collides with this 128 The figure above shows an energy level diagram for the atom hydrogen atom. Several transitions are marked as I, II, III, Momentum ratio ofphotons is . The diagram is only indicative and not to scale. 124 If radiations ofallowed wavelengths from ultraviolet to infrared are passed through hydrogen gas at room temperature, absorption lines will be observed in the nTT (A) Lyman series (B) Balmer series c 3 t.
(C) Both (A) and ( B ) ;

:

. Org'. •
"
II:
.rv:
•:y
:v.i
37
125 In the hydrogen atom, ifthe reference level of potential energy is assumed to be zero at the ground state level. Choose the incorrect statement.
(A) The total energy ofthe shell increases with increase in the valueof«
Figure 1.00
(B) The total energy ofthe shell decrease with increase in the value of n
(A) The transition in which a Balmer series photon absorbed
(Q The difference in total energy of any two shells remains
is vi:
the same
(B) The wavelength oftheradiation involved in transitionll is
P) The total energy at the ground state becomes 13.6eV
486 nm.
126 Choose the correct statement(s) for hydrogen and deuterium atoms (considering the motion ofnucleus) (A) The radius offirst Bohr orbit ofdeuterium is less than that of hydrogen P) The speed ofelectron in first Balmer line of deuterium is more than that of hydrogen (Q The wavelength offirst Balmer line ofdeuterium is more than that of hydrogen P) The angular momentum ofelectron in the first Bohr orbit ofdeuterium is more than that ofhydrogen
127 A neutron collides headon with a stationary hydrogen atom in ground state. Which of the following statements are correct (Assume that the hydrogen atom and neutron has same mass)
(Q IV transition will occur when a hydrogen atom is irradiated with radiation ofwavelength 103nm. p) IV transition will emit the longest wavelength line in the visible portion of the hydrogen spectrum.
129 A hydrogen atomis in the 4^^^ excited state, then: (A) the maximum number ofemitted photons will be 10 p) the maximum number ofemitted photoris will be 6 (Q it can emit three photons in ultraviolet region P) if an infrared photon is generated, then a visible photon may follow this infrared photon 130 An electron with kinetic energy £" eV collides with a hydrogen atom in the ground state. The collision is observed to be elastic for :
(A) 0
=4.94.5 = 0.4eV
=1.5eV
# Illustrative Example 2.4
IfE be the energy of each ejectedphoto electron momentum of electrons is
P= ^2mE
Light ofwavelength 1800 Aejects photoelectrons from aplate ofa metal whose workfunction is 2 eV. If a uniform magnetic
field of5 XlO'^tesla isapplied parallel toplate, what would be Weknowthat changeof momentum is impulse. Herethewhole the radius of the path followed byelectrons ejected normally momentum of electron is gained when it is ejected out thus
from theplatewithmaximum energy.
impulse on surface is given as Solution
J= ijlmE Energy of incident photons in eFis given as
Substituting the values, we get maximumimpulse 12431
J= V2x9.1xI0"^'x0.4xl.6xl0"'^
1800
Aswork function ofmetal is2 eV, themaximum kinetic energyof ^
y = 3.45X10"^^ kg m/sec.
ejected electrons is
# Illustrative Example 2.3 In an experiment tungsten cathode which has a threshold 2300 A
=>
^nm = 6.92eV '^n^=4.9eV
isirradiated byultraviolet lightofwavelength 1800 A.Calculate (i) Maximumenergyof emittedphotoelectron and
If
be the speed of fastest electrons then we have
(ii) Work flmction for tungsten.
4.9 X1.6x1019joule
(Mention both the results in electronvolts) Given Planck's constant./; = 6.6 x 10"^"* joulesec,
2x4.9x1.6x10^®
leV= 1.6 X10~'^joule andvelocity oflightc = 3 x lO^m/sec
9.1x10"^'
=>
Vjj^ = 1.31 XlO^m/s
Solution
The work function of tungsten cathode is
When an electron with this speedentersa uniform magnetic field normallyit follows a circular path whose radius can be given by
,
he
(i)=T— ^th
,
mv r =
qB
[As qvB=^]
I243I
9.1x10"^'xl.31xl0^ r =
1.6x10^^x5x10"^ =>
(i)=5.4eV r=0.149m
Photo Electric Effect & Matter Waves 5
,54
(iii) The energy of incident radiation is 1.89 eV thus the
# Illustrative Example 2.5
corresponding wavelength is The radiation emitted, when an electron jumps from « = 3 to n = l orbit is a hydrogenatom, falls on a metal to producephoto
X=
12431 1.89
•A
electrons. The electrons from the metal surface with maximum
=>
kinetic energy are made to move perpendicularto a magnetic field of(1/320) T in aradius of 10"^ m.Find(i)thekinetic energy
X=6.577.2 A.
# Illustrative Example 2.6
ofelectrons (ii) work function ofmetal and (iii) wavelength of radiation.
Photoelectrons are emitted when 4000 A radiation is incident on a surface of work function 1.9 eV. These photoelectrons
Solution
passthrough a region containing aparticles. A maximum energy electroncombinewith an aparticle to form a He"*" ion, emitting a single photon in this process. He"^ ions thus formed are in
(i) The energy ofelectron in the atom is given by
„
energy level ofhydrogen
their fourth excited state. Find the energies (in eV) of the photons,lying in the 2 to 4 eVrange, that are likelyto be emitted
13.6 ,,
E__ = —r eV
during andafterthe combination. Take/j= 4.14 x 10"'^eVs.
For w= 3, we have „
£3=
13.6
Solution
. ..
^=1.51 eV
(3)'
The energy ofthe incident photon is
and for « = 2, we have
£ =}j^= hp. = 124M Y
£2=^=3.4eV '
Now,
X
=>
(2)2
AE32 = £3£2=151 + 3.4
4000
£. = 3.1 eV
From Einstein's photoelectric equation, the maximum kinetic energy of the emitted electrons is
1.89eV = /iv
Now these radiations incident on a metal surface. The photo electrons emitted from the metal surface move perpendicularly in magnetic field B. Let v be the velocity of photo electrons in a
KEITiuX =/A'd)=3.1eV1.9eV=1.2eV '
It is given that
(electron with £^3^) +He ^ He"'" +photon(in 4*^ excited state).
circle ofradius r. Then,
The fourth excitedstatecorresponds to w= 5. ForHe"^ ion Z=2. Now,we know that the energyof the electron in the state for a hydrogen like ion is
.2 mv
= qvB
mv = qBr
£„=(13.6eV)4
...(2.6)
n
/7 = 5 X10~^kgm/s
Thus,
The kinetic energy of photo electrons is
^
p^ 2m
£, =(13.6eV) x
'
.
(2)2
(5)'
=2.18eV
The energy of the emitted photon in the above combination ICXIU
I
2x(9.1x.10'^31^
reaction is
J
25x10"^®
2x(9.IxlO~^')x(1.6xlO'^)
^ = ^n,ax + (^5) eV
£^»0.86eV (ii) According'to Einstein's photoelectric equation
/iv =(l)+jmvJ^
=>
£=1.2eV + 2.18eV = 3.38eV
This energy is within the range 2 to 4 eV. After the recombination reaction, the electron may undergo
transitions from a higher level to a lower level, thus emitting
photons. Using Equation(2.6), the energies in the lo^r electronic levels ofHe"'' ions are
=>
1.89= (t)+0.86
=>
(t)=1.03eV
'
(13.6eV)> (j)), photoelectrons are ejected and move toward anode with negative polarity.
Intensity/^ >/, Frequency v (same for both radiation)
Incident light v>v.,
^0
Cathode
0
Anode
Reverse voltage Figure 2.10
This voltage V^, we can calculate from equation(2.9) by
substituting v^= 0hence 1
2
yWVmax eVQ=0 V
Figure 2.9
jr I 2 eVQ=^mv^^
Now theelectrons which areejected with very low kinetic energy are attracted back tothecathode because ofitspositive polarity. Those electrons which have high kinetic energies will rush
1
toward, anode and may constitute the current in circuit.
^0=
In this case the fastest electron ejected from cathode will be
^0=
retarded during its journey to anode. As the maximum kinetic 1
2
..(2.10)
/?V(j)
..(2.11)
We cansee onemore thinginfigure2.10 thatthegraphs plotted
7
energyjust after emission at cathode is j ifpotential for two different intensities /j and1^, Vq issame. Current inboth difference across the discharge tube is Kthen the speed v^with the cases incut offatsame reverse potential V^. The reason for this is equation(2.10) and (2.11). If is clearthat the valueof
which electrons will reach anode can be given as I
2
•eV=
•/
...(2.9)
Thus alltheelectrons which are reaching anode will have speed less then or equal to . Remaining electrons which have relatively low kinetic energy will either be attracted to cathode justafter ejection orwillreturn during theirjourneyfrom cathode
depends onlyon the maximum kinetic energy of the ejected electrons which depends only onfrequency oflightandnoton intensity of light. Thus in above two graphs as frequency of incident light is same, the value of
potential difference
is also same. This reverse
at which the fastest photoelectron is
stopped and current in the circuit becomes zero is called cut off
potential or stopping potential.
Photo Eiectric Effect & Matter Waves
58
2.3.4 Effect of Change in Frequency of Light on Stopping Potential
If we repeat the experiment by increasing the frequency of incident light with number of incident photons constant, the variation graph ofcurrent with voltage will beplotted as shown in figure2.11.
Figure 2.12
Frequency (vj > v,)
Hereequation(2.14) can be writtenas
=eFo= /j(vvJ
...(2.16)
This equation(2.16) is called Einstein'sPhotoElectricEffect equation which gives a direct relationship between themaximum kinetic energy stopping potentialfrequency of incident light and the threshold frequency.
Figure 2.11
# Illustrative Example 2.7
This graph is plotted fortwo incident light beams of different
frequency Vj and Vj and having same photon flux. As the Find the frequency oflight which ejectselectronsfrom a metal numberofejected photoelectrons are same in the twocases of incident light here we can see that the pinch offvoltage ^01 as
surfecefully stoppedbya retarding potential of3 V. The photo
well as saturation current
Find the work function for this metal.
are same. But as in the two cases
electric effect begins inthis metal atfrequency of6 x 10'*^ sec"'.
thekinetic energy offastest electron are different asfrequencies are different, the stopping potential for the two cases will be Solution different. In graphIIas frequency ofincident lightis more, the maximumkineticenergyofphotoelectron willalsobehigh and The threshold frequencyfor the given metal surfaceis to stop it high value of stopping potential is needed. These Vjj^ =6xlO"'Hz here
and
can be given as
Thus the work function for metal surface is
...(2.12)
=> and
V •^02 =
hV2
...(2.13)
In general for a given metal withwork function cj), if Vq is the stopping potential for an incident light offrequency v then we
.
(j)=3.978x 1019 j
As stopping potential for the ejected electrons is 3 V, the maximum kinetic energy ofejected electrons will be
have
KE=1>qW
eV^hv^ =>
eVQ =.hvhv^
(t)6.63xlO"3tx6xlo"'
...(2.14)
...(2.15)
KETn3X =3x l.6x 1019J
SJ_ =4.8X10"J According to photo electric effect equation, we have hv = hv,,In +
max
Equation(2.15) shows that stopping potential is linearly proportional to the frequency v ofincidentlight.The variation => frequency ofincident light is (j)Iof stopping potential with frequency v can be shown in figure2.12.
jpfvotoElcctrtcEffect &Matter Waves
59
3.978x10"'^+4.8x10"'^ 6.63x10"^^
V =
According to photo electric effect equation we have
v = 1.32x1015Hz
^max^^l'
# Illustrative Example 2.8
.
^S_=2.851.24eV
^
^max = l61eV
Electrons with maximum kinetic energy 3eV are ejected from a metal surface byultraviolet radiation of wavelength 1500 A. The stopping potential for these ejected electrons can be given Determine the work function of the metal, the threshold wavelength of metal and the stopping potential difference required to stop the emission of electrons.
Solution
as
y _ ^^max
^
„
Vf,=
1.6leV
,
,
= 1.61 volts.
Energy of incident photon in e Fis „
# Illustrative Example 2,10
12431
Determine the Planck's constant h if photoelectrons emitted =>
E:=8.29ev
from a surface of a certain metal by light of frequency
According to photo electriceffect equation, we have E = ^^KE T
=>
^=EKEmax
"
^=8.293eV
=>
(j)=5.29eV
Solution
From photo electric effect equation, we have
Threshold wavelength for the metal surface corresponding to work function 5.29 eV is given as .
Here
/iVj =(f) + eFQj
...(2.17)
and
hv^ = isf + 2eV^
...(2.18)
Subtracting equation(2.17) from equation(2.18), weget
_ 12431 ,
5.29 ^
/i(v2Vi) = e(Fo2Fo,)
=2349.9 A
Stopping potential for the ejected electrons can be given as ^max
and those ejected by light of frequency 4.6 x lo'^ Hz by a reverse potential of 16.5 eV.
max
=>
2.2 X10'^ Hz are fiilly retarded by a reverse potential of6.6 V
3eV
(n2~^oi)(l6xlO^^) h =
(V2V,)
= 3 volt h =
# Illustrative Example 2,9
(16.56.6)(l.6xl0^^) (4.62.2)xl0'^
/z = 6.6x 1034js.
Calculate the velocity ofa photoelectron, ifthe work function ofthe targetmaterial is 1.24eV and the wavelength ofincident
light is4360 A. What retarding potential isnecessary tostop the emission of the electrons ? Solution
Energy of incident photons in e Fon metal surface is „
^
'
12431
.e=2.85eV
•
# Illustrative Example 2,11
When a surface isirradiated with lightofwavelength 4950A, a photo currentappears whichvanishes if a retarding potential greater than 0.6 volt is applied across the photo tube. When a different source of light is used, it is found that the critical retarding potential is changed to 1.1 volt. Find the work function
ofthe emittingsurfaceand the wavelength ofsecondsource. If the photo electrons (after emission from the surface) are subjected to a magnetic fieldof 10tesla, what changeswill be observed in the above two retarding potentials.
Photo Electric Effe^& Matter.Waves j
60
Solution
Solution
In first casethe energy of incident photon in eV is
(a) Let the work function of the surface be If v be the frequency ofthelightfalling onthe surface, thenaccording to Einstein's photoelectric equation, the maximum kinetic energy
12431 •
4950
eV
KE
=2.51 eV
tTlHX
ofemitted electron is given by
The maximum kinetic energyof ejectedelectrons is We know that, = 0.6eV
Thus work function ofmetal surface is given as
=>
\^^ere
T7
^
(j)=2.510.6 eV (l)=1.91eV
=cutoffpotential.
^
In second case the maximum kinetic energy ofejected electrons
A
eK„=^(i> »
eX
AV = V
Now,
^^0
will become
^02
V
^01 he
^niax2^^^02
eX,
£2 = 1.91 + l.leV
EVo=0 e
eXi
e
J X2
Thus the incident energy of photons can be given as
=>
e
e
1_ X^j
X^X2 X.]X.2
...(2.19)
(b) Fromequation(2.19),wehave
Thus the wavelength of incident photons in second case will
he _ AF"o(A.iX.2)
be
c
^ =>
3.01 ^
he
X4129.9A
—X2
10(1.3  0.9)[(4000 X10"'") X(4500x 10"*'®)]
500x10
10
When magnetic field is present there will be no effect on the
stopping potentials asmagnetic force can notchange thekinetic
— =1.44xlO^V/m e
energy of ejected electrons. We use.
# Illustrative Example 2.12
e
cell be reduced from Xj to Xj A, then what will be the change in =>
(b) Light is incident on the cathode of a photocell and the stopping voltages are measured ,for light of two different wavelengths. From the data given below, determine the work function of the metal of the cathode in eV and the value of the universal constant hde.
Wavelength (A)
Stopping voltage (volt)
4000
1.3
4500
• 0.9
eX e
^  h£v
(a) If the wavelength of the light incident on a photoelectric the cutoff potential ?
^0
eX
6
® 4000x10 10
1.3
(j)=2.3eV
# Illustrative Example 2.13
A lowintensity ultraviolet lightofwavelength 2271Airradiates a photocell made ofmolybdenum metal. Ifthe stopping potential is 1.3 V, find the work fimction ofthe metal. Will the photocell work ifit is irradiatedby a high intensity red light ofwavelength 6328 A?
fPhoto'E(6cyjc Effectac MatFer Waves 61
Solution
Theenergy in
f.,
incident photons is Hence, 2271
=>
Which gives T'=2T.
£'=5.47eV
As stopping potential for ejected electrons is1.3 V, the maximum kinetic energyofejectedelectronswill be
It is given that peak emission at temperature T occurs at a
wavelength = 9000 A. If is the wavelength for peak emission at temperature T', then from Wien's displacement law\,T= constant, we have
^
^ _ = 1.3ev
Nowfrom photoelectric effect equation, we have E = ^+KE * max
^'„ =^„!=9000Ak^=3000A. According to the problem, the kinetic energy of the emitted
photoelectrons = excitation energyfor transition «=2to«=3,
=>
i>=E~KE T max
=>
(i)=5.471.3eV
which is
^nw=^32={13.6eV)[^i
(l)=4.17eV
Energy in eV for photons ofred light ofwavelength 6328 A is 6328
,
^™ =13.6'
if)=4.141.89eV
=>
(})=2.25eV
from the surface. After the increase oftemperature the peak radiation from the black body caused photoemission. To bring
Web Reference at www.phvsicsgalaxv.com
these photoelectrons to rest, a potential equivalent to he
Age GroupHigh School Physics  Age 1719 Years
excitation energy between the « = 2 and « = 3 Bohr levels of hydrogen atom isrequired. Findthe workfunction of the metal.
.SectionMODERN PHYSICS Topic  Photoelectric Effect Module Number14 to 18
'

£
due to wavelength
4144A =(4144xl0i0m)
=>
plate
2.425x10'^
4x(3.14)x(2)2 A7=4.82xl0'6m2s'
The maximum kinetic energy of the photoelectrons emitted
from theplate having work ftinction ij) = 1.17 eVis given by
=>
'^max2.581.17
^
^max = l41eV
Themaximum velocity ofphotoelectrons ejected isgiven as
(1.2xl0"^)(4144xl0'^) (6.63xl0"^^)(3xl0^) Kj =0.5x1012
Number ofphotons
y'«vL=1.41eV 2x1.41x1.6x10"'^ 9.1x10"^'
dueto wavelength 4972 A 10(2.4xl0"^)(4972xl0~'") (6.63xl0"^^)(3xl0^)
=>
=>
llktX
= 7.036 X 10^ m/s.
Theradius ofthe circle traversed byphoto electron inmagnetic field B is given by
=
Total photonsA7=«j +«2="0.5 x 10^2 +0.575 x 10'2 =:>
joule
~_Ek
"" {hclk)~ he Number of photons
4.125x10
Number ofphotons striking per squaremeter persecond on the
Energy incident on surfecefor each wavelength in 2 seconds.
The number of photons of wavelength X, are
^
=
/•= 2.424x lO'Vs.
X(1(^ m2)
£ = (1.2 X107) X(2)2.4 X10'joule
—
£71.075x 1012
r =
=>
mv_(9.1xlQ^^)(7.b36xlQ^) (1.6xl0"'^)x(10"'')
mv
,
[AsgvP——]
j=40.0 xl0"^metre=4.0cm.
# Illustrative Example 2.23
a Illustrative Example 2.24
A smallplateofa metal (workfimction =1.17 eV)is placedat a distance of 2 m from a monochromatic light source of
A40 Wultravioletlight source ofwavelength 2480 A illuminate a magnesium {Mg) surface placed 2 m away. Determine the number ofphotons emitted from the source per second and the numberincidenton unit areaof the Mg surfece per second. The photoelectric work function for Mg is 3.68 eV. Calculatethe
wavelength 4800 Aand power 1.0 watt. The light falls normally on the plate. Find the number of photons striking the metal plate per squaremeter per second. If a constantmagnetic field
ofstrength 10^ tesla is applied parallel to the metal surface, findthe radiusofthelargest circular pathfollowed bythe emitted photo electrons.
kinetic energy of the fastest electrons ejectedfrom the surface. Determine themaximum wavelength forwhich thephotoelectric effect can be observed with Mg surface.
Solution
Solution
Energy of incident photon ineVis
The energy of incident photon in eFis given by
r,_ 12431

4800
zr
12431
^=1480"^^
£=2.58eV
=>
£ = 5.01 eF
^
£2.58 X 1.6 X 10'9j
=>
£ = 5.01x1.6x1019;
=>
£=4.125x10'^
=>
£ = 8xl0'9j
Photo Electric
68
The number of photons emitted per second bythe source can be given as
MatterJ/Vayes .
Energy required toeject photoelectron isgiven by (2.22eV)(L60x 10'^J/eV) =3.55x Now exposure time
3.55xlO"'^J
40 N= 8x10
19
t =
= 5x lO'^si
0.1 W
47tm
The number of photons incidenton Mg surface per unit area
x(47ix10"^V^)
/ = 355s.
per second 5x10 N' =
19
4Tcr^
5x10
19
4x3.14x(2)^
7y' = 1.04x lO'^s^m^
12431
(b) Incident photon energy \neV\sE= ^5^ ^^ =>
£:=4.9eV
£ = 4.9x1.6x10'^
The K.E. of the fastest electron is given by ^
KE max =E^T
^
2
KE^ = 5.01 eV 3.68 eV= 1.33eV TTIdX
£=7.84xl0'9j
(c) Atthe cathode (area 4x10"^ m^), the photon flux is
Now the threshold wavelength for Mg surfece can be given as N=
for its work function 3.68 eKis
X"h=^A 3.68
'0.1 W^ 4xlQ^m^ ,47c m^J 7.84x10"'^ J
V=4.06 XlO'^photons/s
X^ =33'1SA
With an efficiencyof 5%, the photocurrent is
1=0.05Ne (0.05) (4.06 x 10'^) (1.60 x 10"'^C)
# Illustrative Example 2.25
7=3.25 X10«A=32.5 nA
A mercury arc lamp provides 0.10 W of UVradiation at a
wavelength ofX= 2537 A (allother wavelengths having been absorbed by filters). The cathode of photoelectric device (a phototube) consists of potassium and has an effective area of
(d) The stopping potential for ejected electrons can be given as
_ ;iv(t)o _ 4.86 eV2.22 eV 0
4 cm^. The anode is located at a distance of 1 m from radiation
source. The work function forpotassiumis (t)^ = 2.22 eV.
=>
e
e
Ko=2.64V
(a) According to classical theory, the radiation from the arc spreads outuniformly in space as spherical wave. Whattimeof # Illustrative Example 2.26 exposureto the radiation should be required for a potassium atom(radius 2 A) in the anodeto accumulate sufficient energy Amonochromatic point source Sradiating wavelength 6000 A, withpower 2 watt, an aperture A of diameter 0.1 m and a large to eject a photoelectron ? screen SOare placed as shown in figure2.17. Aphotoemissive (b) What is the energy of a single photon from the source ? detector D ofsurface area0.5 cm^ is placedat the centreofthe (c) What is the flux of photons (number per second) at the screen (see the figure)." The efficiency of the detector for the cathode ? To what saturation current does this flux correspond photoelectron generation per incident photon is 0.9. if the photoconversion efficiency is 5% (i.e., if each photon (a) Calculate the photon flux at the centre of the screen and has a probability of 0.05 of ejecting an electron). (d) What is the cut off potential
the photocurrent in the detector. SC
? A
1,
Solution S.
(a) The UV energy flux at a distance ofone metre =
0.1
'.J
D
1 \
wattper square metre.
Crosssectional area of atom A = tzP'
=>
\ /
.4= 7c(2x 10~'®m)^ = 4;rx lO20m^
6 m
Figure 2.17
•»
; Phdto Efectric Effect
Matter Waves
69
(b) If a concave lens L of focal length 0.6 m is inserted in the aperture as shown, find the new values of photon flux and photocurrent. Assume uniform average transmission of 80%
Thus the concave lens forms its image at a distance//2 i.e., 0.3 metre on the left of the lens. This we get from lens formula
1 = 1_1
from the lens.
/
(c) Ifthe work function ofthephotoemissive surface is 1 eV, calculate the value, of the stopping potential in the two cases (without and with the lens in the aperture).
V
u
1 = 1+1 = ^ + ^ V
f
u
0.6
0.6
1
0.3
v=0.3m Solution
Now all the photons are focussed at a distance 0.3 m from the lens or 5.7 m from the detector. The detector transmits 80% of
(a) Energy of Photon is
^ he 6.63x10"^^ x(3xl0^)
photons. So the number of photons reaching the detector 80
n'=
~ ~ 6000x10^'® £ = 3.315 X10'>ule
Solid angle subtended by detector Solid anglesubtended by aperture
Number ofphotons emitted per second from the source are
80
Power n
=
n
=
n'= TrtTT Xphoton flux x
^
(0.5x10"*/5.7)^
Tlx (0.05)2z 7(0.3)27
Substitute the value of photon flux and obtain the value of
2
=>
100
Energy of each photon 3.315x10
Xphoton flux
19
« = 6.033 x lO'^photorisec,l
Here the solid angle subtended by detector is less than that subtended by aperture on the source. This shows that all the photons reaching the detector are allowed by aperture. In this
Photon flux
(bp =
^P =
=>
,—
area of detector
0.5x10
4
per m^ per second
({)p =2.956 X10'Vm^ sec.
way the number of photons reaching the detector per m^ per Thus photon current = 0.9 x Photon flux x e
second is given by Photon flux
(bp =
7 6.033x10
18
4x(3.14)x(6.0)^ =>
=>
/= 0.9XPhoton flux X1.6x 10"'^Amp.
=>
7=0.0213 pA.
m 2cl
(j)p = 1.33 XlO'^photons/m^sec
(c) The stopping potential is independent ofphotocurrent. In both the cases, according to Einstein's photoelectric equation, this is given by
Photo current = ne
E= W+eV^ 1= [9.0 Xphoton flux x area ofdetector] e
=>
/=[0.9 XphotonfluxX(0.5 x 10^)] x 1.6x 10"'^ Amp.
=>
/=0.096 pA.
=>
3.315 X 10"'^ = leK+eK 3.315x10
19
1.6x10"*^
(b) The source is at the focus of concave lens.
eV=\eV+eV.
2MeV=\eV+eV^ K^=2.06l K= 1.06 volt.
Web Reference at www.phvsicsgalaxv.com
Age Group  High School Physics  Age 1719 Years
0.3 m
Section  MODERN PHYSICS r«0.6 mW 5.7 m
Topic Wave Particle Duality ModuleNumber  3 to 11
Figure 2.18
Photo Electric Effect &Matter Waves •
70
Practice Exercise 2.3
(vii)
When the sun is directly overhead, the surface of the
earth receives 1.4 x 10^ W/m^ ofsunlight. Assume that the light
(i) Calculate the number ofphotons emitted per second by ismonochromatic with average wavelength 5000 Aand that no a 10 watt sodium vapour lamp. Assume that 60% of the light is absorbed in between the sun and the earth's surface. consumed energy is converted into light. Wavelength of the The distance between the sun and the earth is 1.5 x 10" m. (a) Calculate the number ofthe photons falling per second sodium light = 5900 A. on each quare meter of earth's surfacedirectlybelowthe sun. [1.7 X 10"] (b) How many photons are there in each cubic meter near
(ii)
the earth's surface at any instant ?
Abulb lamp emits light ofmean wavelength of4500 A (c)
The lamp is rated at 150 watt and 8% of the energy appears as emitted light. Howmany photons are emitted bythe lamp per
How manyphotons does the sun emits per second ?
[3.5 X lO^', 1.2 X 10'3, 9.9 X 10''^]
second.
[27.17 X 10"!
2.6 Wave Particle Duality
(iii)
The true nature of light is very difficult to asses. In 1801Young's double slit experiment it was shown that light exhibit fundamentalpropertieslike a wavediffraction and interference. After several years Maxwell discovered that light was a wave
0.05 kg of ice at  20°C is to be converted into steam at
100°C. The ice is first heated with a420Wattsheaterfor5 minutes.
After this, itisheated with aninfi^ared lamp ofX, = 10,000 Afor 23 minutes and 20 seconds at an efficiencyof50%. Find the rate
at which the photons are striking the ice when heated with infi:aredlamp.
Specific heat of iceat20''C = 500cal/kg
Specific heatofwater= 10^ cal/kg
of oscillatingelectricand magnetic fields, an electromagnetic wave. On the other hand Plank's equation theory, photoelectric effect and campton effects indicates that light behaves like a particle, a photon, withbothenergyand momentum. Thus light shows wave particle duality.
Latent heat offusion ofice = 8 x 10"* cal/kg
Latentheatofvaporization ofwater= 5.42x 10^ cal/kg.
2.6.1 Momentum of a Photon
[2 X 102%ec']
According to relativistic theorythe total relativistic energy£" of a particle is related to the magnitudep of the momentum and
(iv)
Alight source, emitting three wavelengths 5000 A, 6000 A
mass m
and 7000A hasa totalpower of 10"^ Watt and a beam diameter of 0.002 m. The power density is distributed equally amongst the three wavelengths.The beam shines normally on a metallic surface of area 10"^ m^ and having a work function of 1.9 eV. Assuming that each photon liberates an electron, calculate the charge emitted per unit area in one second.
as
£:2^^2^2 + ^2^4 For a photon, its mass is zero, thus we have E=pc E p = — ^ c
[93.76 coul/mVsec]
For a photon its energy can be given as
(v) A monochromatic light sourceofintensity5 mW emits. 8 X 10^^ photons per second. This light ejects photoelectrons
Thus photon momentum is given as
fiom a metal surface. The stopping potential for this setup is 2.0 eV, calculate the work fiinction ofthe metal. [1.9 eV]
(vi) A potassium surface is placed 75 cmawayffoma lOOW bulb. It is found that the energy radiated by the bulb is 5% of the input power. Consider each potassium atom as a circular disc of diameter 1 A & determine the time required for each atom to absorb an amount ofenergy equal its work function of 2.0 eV.What is the answer ifthe atom is assumed to be spherical. [57.6 s]
...(2.35)
Pphoton
^

Q
/'photon^ X
...(2.36)
This relation in equation(2.36) gives the momentum ofeach photon of an electromagnetic wave having a wavelength X.
2.7 DeBroglie's Hypothesis In year 1923, as a graduate student Louis deBroglie made a surprising suggestion which later confirmed in 1927 by different experiments. By the year 1921 it was all accepted that light
£hoto Efefric Effect &Matter Waves behaves like a particle, a stream ofphotons andthewavelength
nh
of light is related to the magnitude of its momentum as x=
h
...(2.37)
/'photon
Using the above accepted fact, Broglie started thinking the reverse ofit. He suggestedthat since light waves could exhibit particle like behaviour, particles ofmatter should exhibit wave
mvr =
...(2.42)
2n
Thus if a wavelength is associated with the electron, the quantization ofatomic orbitals leads toconsider integral number ofwavelengths that fit intoeachorbit Forexample figure2.19 shows the different orbits in which for different values of n
electron wave isforming astationarywave. Figure2.19(a) shows likebehaviour. DeBroglie proposed thatall moving mattCT has an orbit inwhich thestationary wave isformed fortheprinciple
.a wavelength associated with it, just as a wave does and this wavelength is related to the magnitude of its momentum by an equation of the same form as given in equation(2.37) as
quantum number « = 3. Similarlyfigure2.19(b) shows an orbit correspondingto « = 4 and figure2.19(c) shows an orbitwhich is larger then that of«=3 but smaller then that of« = 4 in which
we can see that in this orbit wavelength is not fit properly thus X=——
...(2.38)
/'particle
The wavelength associated with a particleis calledits deBroglie wavelength. The deBroglie hypothesis implies that the wave particle duality has a universal and symmetrical character'i.e. waveshave particle properties and moving particles have wave properties. It was also assumed that as the speed ofa particle increases its wave character also increases and the wavelength associated with the particle may have a significant practical values. For particles in nature moving very slowly, the wave character associated with the particle will not have much significance.
standing waves can not be formed for this orbit hence the orbit is not stable.
orbit
fern = 3
(a)
2.7.1 Explanation ofBohr's Second Postulate
The deBroglie hypothesis brought Bohr's second postulate in a new light. It was stated that the magnitude of angular momentum ofthe orbiting electron in a hydrogenic atom was an integral multiple ofh/2n. Given as ...(2.39)
orbit
fern = 4
Asin anorbit speed ofelectron is oftheorder of10^ m/sitmay have a significant wave character associated with it. As we know in an atom stable orbits are those in which electron does
not radiate any electromagnetic radiation. Thus ifwe describe electron as a wave, its stable orbits in an atom are those that
(b)
satisfy the conditions for a stationary wave so that the complete wave energy is sustained in that orbit only.Thus for stationary waves to exist in an orbit the circumference ofthe orbit must be
an integral multiple ofthe wavelength, which can be given for an orbit of radius r
...(2.40)
orbit
For electron, its deBroglie wavelength \ can be given as for 3 < n < 4
^
mv
...(2.41)
Now from equation(2.41) & (2.42) we have (c) 2w = n — mv
Figure 2.19
Pholo''ae^trid iffedt: rM^er VVav^j
si
a_ = 0.7
2.8 Radiation Pressure
a.0.3
We've discussed that a,photon carries mornentum, given as .
, P
...(2.43)
Pwatt
When a photon beam incidents on a surface,andabsorbed, the
Figure 2.21
total momentum can^ied by the beam is also transferred to that surface hence a force is exerted on the surface. Ifthe surface on
whichthe^beam incidentis reflectingthen dueto the reflection in.photon beamthe change in momentum,of photons will be twice the value compared to the.case when it was absorbed. Thus 'force exerted on surface will also get almost doubled. Now we'll discuss different cases ofincidence ofa light beam
In'this case 70% of the incident photons are reflected back and 30% are absorbed by the'body. Thus the photon which is absorbedwill impart a momentum
r,' OJFX
X
''he •
X
to
h'
X .—^
he ' ' X
. ,..(2.48)
F=
...(2.44)
he
2h 1, 0.3P?.
I.IF
2.8.1 Force Exerted by a Light Beam on a Surface
Figure2.20 shows a black body ofmass m placed on a smooth surfeceon which a light beam of crosssectional area S incident. The beam is produced by a torch of power P watt. If X is the wavelength oflight produced by torch, the number ofphotons emitted per second are
'2>fi
the body. Thus net force acting on body can be given as F=
FX
'
to the bodyapd the photon.
which is reflected will impart the change in momentum
on different surfaces.
N=
h
If in any of the case we wish to find pressure on the surface on which light beam is incident.In aboverelationswe can use.light
intensity
% i^st^ad ofpower Pofthe light source;
2.8.2 Force Exerted on any Object in tiie Path of a Light Beam
Figure2.22 shows a big lamp of power F watt which produces a uniform parallel beam oflight ofcross sectional area S. Thus the intensity ofthis light beam will be '
Black Body
/= ^ w/m^•
...(2.49)
P watt
• r """M
area = S
1
1
Figure 2.20
We know that momentum in each photon is
h
Power = P watt
Area =
P=x
'
Area = 5
....(2.45)
(a)
As all the photons incident on the black body will be absorbed by it, here the total momentum absorbed by the body per second or force exerted on the body is F=
FX he
P
...(2.46)
c
In above case ifthe surface ofbody is perfectly reflecting like a Power = P watt
mirror then the force exerted on body will become Area = nR? Area = S
...(2.47) ,
c
Figure 2.22
Similarly consider another case as shown in figure2.21. Now
the surface ofbodyon which light beam is incident is having a jf jj, reflection coefficient
0.7 and absorption coefficient a^=0.3.
path ofthis beam a black body sphere is placed as
shown infigure2.22(a). Inthiscase only those photons will be
,Ph^o EJectric Effect &Matter Waves 73''
incident on the sphere which pass through the cross sectional areaA=idf which isthe projection ofsphere on acrosssectional plane. Thus the power incident on sphere is
Inthe path ofauniform light beam oflarge crosssectional area
P = IA=I%R^
...(2.50)
Thus forceexertedon sphere will be
C
C
Illustrative Example 2.27
and intensity I, a solid sphere ofradius R which is perfectly reflecting isplaced. Find the force exerted on this sphere due to the light beam.
...(2.51)
'
Solution
Similarlyasshown infigure2.22(b) a black body cone isplaced in the path ofthe lightbeam. Here alsothe projection of cone along a crosssectional plane of beam is^ =71/?^ hence the force
Light Intensity /
exerted on cone due to the beam will remain same as F =
InR
...(2.52)
2.8.3 Force Exerted bya LightBeamat Oblique Incidence / dF%mQ
As shown in figure2.23 whena lightbeamincidenton a mirror
at an angle 0 to normal, it will be reflected at same angle. If power of lightbeam isP watt, themomentum ofphotons inthe beam per second will be ...(2.53)
Figure 2.24
To find theforce onsphere, we consider asmall elemental strip ofangular width onitssurface atanangle 0from its horizontal diameter as shown. TheareadSof this stripon the surface of sphere is
Ap sin 9
1I
dS = 2nR sin0 • RdQ
.*..(155)
Nowas shown in figure2.24, dA is the projection of the slant strip areadSalong the crosssectional plane of thelightbeam
Reflecting Surface
and it is given as
dA = dSeos 0
...(2.56)
Ap cos 9
f.
Here thepower oflightincident on thisstrip is
>F
Apcos9
'
dP=ldA
Thus the momentum of photons per second incident on this strip are ,
dP
IdA
dp= — = ~
...(2.57)
Here wecan see thatthese photons areincident atan angle 0to
v.
tip sin 9
the normal
ofthis strip and as sphere isperfectly reflecting.
These are reflected at the same angle 0 to Pwatt
as shown in
figure2.24. Figure 2.23
Here as shown in figure2.23 there will be no change in the component of this momentum along the mirror due to reflection
butalong normal momentum oflightbeam is changed byItsp cos0. Thus force exerted on themirror isalongitsnormaland is given by IP
F= 2 Ap cos 0 = — cos c
...(2.54)
Herethe changein momentum ofphotons is alongthe normal andthus force exerted on this stripalongthe normal is dF = 2dp cos0 ^ —— cos0 Thus net force on sphere will be given as
F=Ji^FcosG ^pi dA_^2
F=
cos
0
... (2.58)
Photo "Electric Eff^t &Matter VVaves ]
74
Now to find the force on the cone, we consider an elemental
strip ofwidth dx on the lateral surface ofcone ata distance x
F= J—(27ii! sin 0cos0• cos
from the vertex O ofconeas shown in figure2.26. If the radius of thestripis r its surface areais
0
2^
iLE^— [cos^ BsinBt/G c
...(2.60)
dS = 2Tir • dx
J 0
41 nR'
Herebysimilartriangles in conewehave
cos 0
F=
^ IizR'
F=
R
^ [10]
^1r^+h^
=>
InR'
... (2.59)
F=
r = X sin 0
Thus area of strip can be given as
We can see that equation(2.58) is exactly same as equation
(2.59). Thus for asphere placed in the path ofa light beam, force exerted on sphere isindependent from thenature ofthesurface ofsphere. Butthis happens onlyfor a sphere.
...(2.61)
dS = 27a: sin 0 •
IfdA betheprojection ofslant strip ofarea dSalong thecrosssectional planeof the lightbeam, it can be given as sin 0
••(2.62)
Lets discuss the same for a cone with reflecting surface in the
IfdPisthe power oflight beam incident on the strip then
path ofa light beam, illustrated innext example.
dP=IdA
Illustrative Example 2.28
...(2.63)
Now themomentum persecond ofthelightphotons dpincident Figure2.25 shows acone ofradius Rand heightwith perfectly on the strip is reflecting lateral surface, isplaced inthe path ofalight beam of _ dP IdA ...(2.64) intensity/. Find the force exerted on this cone.
^
c
c
Here also we can see that from figure2.26 the momentum of
photons is changing only along normal due to reflection of these photons, which will exert a force on cone along normal
Intensity = / watt/m^
direction. Ifthis force is dF, it can be given as ...(2.65)
dF=2dPsmQ
Thus net force on cone can be given as
Figure 2.25
sin 0 Solution
=>
F=J'2(^sin^0 F
sin^ 0
rfHicosiS
^£/5sin^0
v!/ oFsm0
F
•27ad!xsin 0
4/7tsin 0 F=
Figure 2.26
iR
^xdx 0
iphoto Electric Effect & Matter Waves
75
2.8.5 Variation in Wavelength ofEmitted Photon with State of +H'
F
r =
Motion of an Atom
4/71 sm^ . 4I c
As we've discussed that when a photon is emitted by transition
ofan electron in an atom,the atomrecoils and therecoil velocity can be obtainedbymomentumconservation. In previoussection we've discussed that the emitted photon wavelength is given
R
4{r^+h^)
by Rydberg's formula as 1
[As sin 0 =
21 kR
4{R^+H^) ...(2.66)
F=
1
...(2.70)
R
Equation(2.66). gives the force exerted by light beam on a perfectly reflecting cone in its path. 2.8.4 Recoiling of an Atom Due to Electron Transition We know that when an e~ in a hydrogenic atom makes a transition from a higher energy level to a lower energy level
«p thedifference ofenergies ofthe two energy level isreleased in the form ofan electromagnetic radiation photon. As we've discussed that every photon has a momentum associated with it, we can say that when a photon is emitted from an atom, it
This equation(2.70) or Rydbergformula is valid only ifall the energy released by electron during transition from energy level
«2 to«j will transform into thephoton energy. Butin ourcase if all the energy released by electron is carried by the photon then no energy is left with which the atom an recoil. If it does not recoil, the law of momentum conservation will violet. Thus in
previous section equation(2.69) only gives an approximate value ofrecoil velocity. Practically we can say that the energy released by electron during transition is shared between the emitted photon and the kinetic energy ofrecoiled atom. IfX'is the wavelength ofemitted photon and is the recoil velocity of atom, we can write the energy equation as
must recoil in opposite direction as shown in figure2.27 Rch2^
1
1
he
r
+ ~mvi
...(2.71)
Using momentum conservation we can write
r =Mv„ Hatom
Photon
Figure 2.27
Here the wavelength of emitted photon can be given as 1
1
...(2.67)
The momentum ofthis emitted photon can be given as
p=^=hRZ:^
1
1
...(2.68)
If the after emission ofphoton the atom recoils with speed v^, using momentum conservation principle, we can write
...(2.72)
Solving equation(2.71) and (2.72), we can calculate the exact value ofrecoil velocity and the wavelength ofemitted photon X'. Note that the value of will be slightly more compared to that calculated in previous section by equation(2.67). As the difference between the two is very small, for numerical applications students can use equation(2.69) and (2.70) for calculations of emitted wavelength and recoil velocity if very high accuracy is not needed in the application. 2.8.6 Variation in Wavelength of Photon During Reflection We've discussed that when light incident on a body a force is exerted due to the momentum associated with the light photons. Consider the following example. A torch of power P in front of a body ofmass Mis lit just for a time At so that torch will emit a pulse of energy AE which is given as AE=PAt
...(2.73)
Ifwavelength light is X, the momentum in this pulse is hRZ'^ v„ =
M
h _0
...(2.69) AP =
AE
(.YD
h X
AE
...(2.74)
Photo Electric,Effect^^ Matter Waves ;
76
As the total energy ofsystem isalso constant, the equation for A£
energyconservation can be given as ...(2.78)
Power = P watt
Ap
Now solving equation(2.77) and (2.78) we can get the value of speed ofblock as well as the wavelength X' ofreflected light pulse. This process gives accurate results but for numerical problems, itis relatively lengthyprocedure. Students are advised to directly use equation(2.75) or assume that the reflected wavelength is approximately same because the difference is very small which can be neglected for numerical problems.
(a)
Power = Pwatt
(b)
Figure 2.28
# Illustrative Example 2.29
Asshown in figure2.28 we can see thatwhen theenergy pulse is reflected from the body if its surface is perfectly reflecting. Due to this the body will gain a velocity v as shown in
With whatvelocitymustan electron travel sothatitsmomentum
is equal to that ofphoton with awavelength of?. =5200 A.
figure2.28(b). Solution
Hereif we assumethat the wavelength ofreflected and incident
light are same, we can say that themomentum ofreflected pulse is same in magnitude but direction is opposite. Thus by
Momentum of photon 6.626x10
momentum conservation we can write
^
Ap = Mv  Ap
kg.m/s.
Where v is the velocity of electron.
... (2.75)
Using the above equation we can find the speed of theblock after reflecting the pulse. But this result will only be an approximate value. Ifwe carefully look into thesituation once again, aquestion arises, ifwhole ofthe energy ofpulse isreflected
back, how the block has gained the kinetic energy
5200x10""^
Momentum ofelectron = wv = 9.1 x 10"^'v
2 tsp=Mv =Mv
X
34
Equating the two momenta, wehave 9.1 X10^'v=6.626x l(r3''/5200x 1010 >
v = 1400m/s.
# Illustrative Example 2.30
MvP. Hydrogen gas in theatomic state is excited toan energy level
Thus it is sure that when the incident energy pulse is reflected
from the block, someof its energyis transmitted toblock'as its
such that the electrostaticpotentialenergyof Hatombecomes  1.7 eV. Now the photoelectric plate having work function
kinetic energy and remaining is reflected as energy pulse of () = 2.3 eV is exposed to the emission spectra of this gas. energy A£". Now the way ofaccurate calculations for the speed Assuming allthetransitions tobepossible, find theminimum deBrbglie wavelength of ejected photoelectrons. ofblock is given here. The number of photons in the incidentpulse are
Solution
AE
N=
Given that electrostatic potential energy of Hatom is
i'Yx)
PE=\JqV N=
AEX he
... (2.76) . We know that kinetic energy is given as PE
Here Xis the wavelength ofincident light.
=^=0.85eV
If?.' is the wavelength ofreflected light, the equation for Thus total energy momentum conservation can be written as
N^=Mv~N^
p=_I.7 + 0.85=0.85eV
•••(277)
13.6 £
= —
= 0.85 eV
[Photo^ Electric Eff^ &Matter Waves 2_ 136
n' =
0.85
77
= 16
and
n=A
_
Aa 2 • •
Hence the atom isexcited to state n=4.The maximum energy is emitted when electrons will make a transition fiom w= 4 to « = I for which energy emitted is
# Illustrative Example 2.32
Find the ratio ofdeBroglie wavelength ofproton and aparticle
AE= 0.85(13.6) = 12.75eV
Now this photon energy when incident on ametal plate having work function 2.3 eV, the kinetic energy offastest electron ejected
which have been accelerated through same potential difference. Solution
can be given as
Kinetic energy gained by a charge q after being accelerated
^max =^'1'= 12.752.3 = 10.45 eV
through a potential difference Fvolt,
Theminimum deBroglie wavelength is given by .j
_
h
qV= ^m^P•
h
(/')max " pm(K.E.)^
V=J^ m
=>
6.63x10^^^
V2x(9.].1x103')(10.45x1.6x10'9) =>
mv= ^2mqV
Now we have, deBroglie wavelength is given as h
^^ = 3.8xlOiOn]3.8A
_
h
jlmqV
# Illustrative Example 2.31
fripqpVp Anaparticle anda proton arefired through thesame magnetic fields which is perpendicular to their velocity vectors. The a
Substituting V^=V
particle and the proton move such that radius of curvature of
4x2
their pathissame. Findtheratio oftheirdeBroglie wavelengths. Solution
# Illustrative Example 2.33
Magnetic force experienced bya charged particle ina magnetic field is given by,
canform a standing wave between theatoms arranged in a one
Eg=q'^y.B =qvBsmQ In our case,
Hence,
= qvB
[As
mv
Bqv = ^
=>
Assume that the deBroglie wave associated with an electron
mv = qBr
0 = 90°]
dimensional array with nodes at each of the atomic sites. It is found that one such standing wave is formed if the distanced
between theatoms ofthearray is2 A. Asimilar standing wave is again formed if d is increase to 2.5 A but not for any intermediate value of d. Find he energy of the electrons in electron volt and the least value ofd for which the standing wave of the type described above can forms.
The deBrogliewavelength
X= ~mv
^ qBr
Solution
From the given .situation it is clear that stationary waves are ^apanicle 5i'proton
Since,
'
^=1
formed for two successive values of2 A and 2.5 A thus we can directly say that
y=0.5A ;^=2x0.5 = 1A
Photo Electric Effect & Matter Waves
78
Thus the energy of electrons can be given as E =
he
®
(6.63xl0^'^)(3xI0^)
E
(5x1.6x10")
2m
?.= 2.48625 X10"^ m= 2486A
(h/xy E =
E =
2m
(6.63x10"^'^)^ (10"'®)^x2x9.1xl0^^
h deBroglie
h
p
£• = 2.41 X lOi'^J 2.41x10 £ =
1.6x10
6.63x10 deBroglie
17
19
•eV
34
^2x(9.1xl0^^)x(2xl.6xl0'^)
Wie=8.6877A
£• = 150.6 eV
Now we use
Forformation of stationary waves between two atomic sitesthe X
minimum separation must be that is0.5 A. # Illustrative Example 2.34
X
2486
^deBroglie
8.6877
=286.18
(c) After time t, thepotential Vofthe sphere is given by
In a photo electric effect setup, a pointsource light ofpower 3.2 X 10"^ W emits monoenergetic photons of energy 5.0 eV.
47teo yr
The source is located at a distance of0.8 m from the centre ofa
where q = charge on the sphere.
stationary metallic sphere ofwork function 3.0eV andofradius 8.0 X10"^ m. The efficiency of photoelectron emissionis one
We also use stopping potential as
for every 10^ incident photons. Assume that the sphere is
^ ^photon "*1*
located and initially neutral, and that photoelectrons are instantly swept away after emission.
When potential ofsphere becomes 2volt, photo electric emission (a) Calculate the number of photoelectrons emitted per stops as maximum kinetic energy ofpTiotoelectrons is2eV. second.
(b) Find the ratio of the wavelength of incident light to the DeBroglie wavelength ofthe fastest photoelectrons emitted. (d) (c) It is observed that the photoelectrons emission stops at a
(9xl0^)«(1.6xl0^^) ^
certain time t after the light source is switchedon. Why ? (d) Evaluatethe time t.
8x10"^ Solving we get
Solution
n = l.llxi07
(a) The energy reachingthe sphereis givenby E'=
P
2
. XTtr^ =
Now, we use
Pr^
« 10^
t = —r
l.llxio'
TTlcUv
'
P)
mdx
is doubled butn remains the same
228 The frequency and the photon flux of a beam of light falling onthesurface ofphotoelectric material are increased by 234 A metal surfaceis illuminatedbya light ofgiven intensity a factor oftwo; This will and frequency to cause photoemission. If the intensity of (A) Increase themaximum kinetic energy ofthephotoelectrons, illumination is reduced to one fourth ofits original value keeping as wellas photoelectric currentbya factor of two frequency ofradtion constant, then the maximum kinetic energy (B) Increase the maximum kinetic energy ofthe photo electrons of the emitted photoelectrons would become : andwould increase thephoto electric current bya factor oftwo (A) Unchanged P) l/16thoforiginalvalue (Q Increase the maximum kinetic energy ofthephoto electrons (Q Twice the original value p) Four times the original value by a factor oftwo and will have no effect on the magnitude of thephoto electrons by a factor oftwo and will have noeffect on 235 A point source causes photoelectric effect from a small the magnitude ofthe photoelectric current produced P) Noproduce any effect onthekinetic energy oftheemitted metal plate. Which ofthe following curves may represent the electrons but will increasethe photo electric currentbya factor of two.
229 Let
saturation photocurrent as a function of the distance between
the source and the metal?
be the maximum kinetic energy ofphotoelectrons
emitted by light of wavelength X., and
corresponding to
wavelength IfA.j = 2X^ then : (A) 2K^=K^ P) K^=2K^ P) ^,> 2^:2 (Q K^.): (A)
nh iTzXm
(B)
Xm
Xm Xm
Innh
(Q
21!L
P) increase themaximum kinetic energy ofthephotoelectrons bya factor greater thantwoandwould increase thephotoelectric saturation current by a factor of two.
(Q increase the maximum kinetic energy ofthephotoelectrons by a factor greater than two and will have no effect on the magnitude ofthe photoelectric saturation current produced. P) increase themaximum kinetic energy oftheemitted photo electrons by a factor of two but will have no effect.on the saturation photoelectric current.
P)
211 Minimumlight intensitythat can beperceived bynormal humaneye is about 10"'® Wm2. What is theminimum number
25 No photoelectrons are emitted from a metal if the wavelength oflightexceeds 6000 A.Thework function ofthe ofphotons ofwavelength 660 nm that must enter the pupil in one second,for one to see the object? Area of crosssection of metal is approximately equal to: thepupil is 10"^ m2? (A) 3.315xl(r6j P) 3.315 xl0'9j
(Q 2.07x10'^ J
p) 2.07 X1022 J
(A) 3.318 xlP (Q 3.318 xlO^
P) 1.453x103 p) 1.453x105
26 5% of the energy supplied to a lamp is radiated as a visible
light. How many quanta of light are emitted per second by 212 The average wavelength of deBroglie wave associated 100wattlamp.Assumethe average wavelength of visiblelight with a thermal neutron ofmass m at absolute temperature T is given by (here k is the Boltzmann constant):
as555nm?
(A) 0.75x10'^ (C) 2.16 X10^^
P) 1.39x10'^ P) 2.83 X10'^
h
(A)
•JmkT h
27 All electrons ejected from a surface by incident light of
(Q
yl3mkT
_
P)
h
•^2mkT h
P)
lylmkT
wavelength 2000A canbestopped before travelling 1m in the direction ofuniform electric field of4 N/C. The work function of the surface is:
(A) 4eV (Q 2eV
P) 6.2eV P) 2.2 eV
.
213 What percentage increase in wavelength leads to 75% loss of photon energy in a photonelectron collision? (A) 200% P) 100% (Q 67.7% P) 300%
[photo Elecfric Effect &Matter Waves
214 Aparallel beam oflight ofintensity7and cross section
219 The work function ofasubstance is 4.0 eV. The longest
area S is incident on a plate at normal incidence. The
wavelength oflightthatcancause photoelectron emission fiom this substance is approximately: isVand the work function ofthe plate is(p (hv > 3) & 2.36 eV (4 > 3)]
236 At a given instantthereare 25%undecayed radioactive
nuclei in a sample. After 10 second thenumber ofundecayed nuclei reduces to 12.5%. Calculate: (0 Mean life ofthe nuclei
is one for every 10^ incident photons. Assume that the sphere (ii) The time in which the number of undecayed nuclei will
is isolated and initially neutral, and that photo electrons are instantly swept away after emission.
further reduce to 6.25% ofthe reduced number. Ans. [(i) 14.43 s, (ii) 40 s]
Photo Electric Effect & Matte^Vyay^_^
: 98___
237 When a beam of10.6 eV photons ofintensity 2.0 W/m^ 241 Irradiating themetal surface successively by radiations fells on aplatinum surfece ofarea 1.0 10"^ m^ and work function ofwavelength 3000 Aand 5400 A, itisfound that the maximum 5.6 eV, 0.53% ofthe incident photons eject photoelectrons. Find velocities ofelectrons areintheratio2:1. Findtheworkfiinction the number of photoelectrons emitted per second and their ofthe metal surface, meV.
minimum energies (in eF)Take 1eV= 1.6 x10"'^ J. Ans. [6.25 >= lO", 0]
238 In a photoelectric effect experiment, photons of energy 5 eV are incident on the photocathode of work function 3 eV.
Ans. [1690]
242 A filter transmits onlytheradiation ofwavelength greater than 4400 A. Radiation from a hydrogen discharge tube goes
through such a filter and isincident on ametal ofwork function
= lO'^ m"^ s~\ saturation current of 2.0 eV. Find the stopping potential which can stop the
For photon intensity
4.0 pA is obtained. Sketch the variation of photocurrent against the anode voltage in the figure below for photon
photoelectrons. Ans. [0.55 volts]
intensity (curve A) and= 2 x lO'^ m"^ s"' (curveB). 243 The peak emission from a black body at a certain
temperature occurs at awavelength of9000 A. On increasing the temperature, the total radiation emitted isincreased 81 times. Atthe initial temperature when the peak radiation from theblack body is incident on a metal surface, it does not cause any Ans. [ 6^2
0
2
4
6
V/yol. k)
239 In a photoelectric setup, the radiations from theBalmer series ofhydrogen atom are incident on a metal surface ofwork
photoemission from the surfece. After the increase oftemperature the peak radiation from the black body caused photoemission. To bring these photoelectrons torest, a potential equivalent to theexcitation energy between the«=2 and «= 3 Bohr levels of hydrogen atom isrequired. Find the work function ofthe metal. Ans. [2.25 eV]
function 2 eV. Thewavelength ofincident radiations liesbetween 450 nm to 700 nm. Find the maximum kinetic energy of
244 The radiation emitted when an electron jumps from n = 3
photoelectron emitted. (Given hc/e = 1242 eVnm).
ton = 2 orbit of hydrogen atom falls on a metal to produce
photoelectrons. The electrons emitted from the metal surface 240 Monochromatic radiations ofX= 640.2 nm from a neon
withmaximum kinetic energy aremade to moveperpendicular
. . lamp irradiates photosensitive material made of cesium on toa magnetic field of 1 T ina radius of10 m. Find tungsten. Thestopping potential is measured as 0.54 V. The source is replaced by an iron source and its 427.2 nm line (a) the kinetic energy of the electrons, irradiates thesame photocell. Predict thedifference in stopping (b) theworkfunctionofthemetaland
potential. (Round offto singledigit) Ans. [1 eVJ
(c) the wavelength ofthe radiation. • (Planck's constanth = 6.62 x Ans. [(a) 0.86 eV, (b) 1.03 eV, (c) 6570 A]
Js)
3
XRays FEW WORDS FOR STUDENTS
We have already discussed about the energy levels in atoms. In
this chapter we will study about the utilization of the energy releasedfrom atomic transitions in heavy atoms in theform of Xrays. Now we will study from the discovery ofXrays to the fundamental utility along with the different production
Metal
target
___ .HighVsr voltage
X rays
source
mechanism ofXrays, Vacuum
Electrons Heated filament
CHAPTER CONTENTS
3.1
Introduction toXRays
3.2
Production Mechanism ofXrays
3.3
Moseley'sLaw
3.4
Applications ofXrays
Filament
voltage
COVER APPLICATION
Figure(a)
Figure(b)
Figure(a) shows a commercial XRay machine used in various diagnostic labs in which Xrays are used^to take photographs of
organs of human body. Figure{b) is a typical XRay image of human hands taken on a commercial XRay machine.
,X:RaysJ ;ioo
3.1 Introduction to XRays
(d) The wave nature ofXrays makes them useful tool for the study ofthe structure ofcrystals &molecules. This was later
The discovery of Xrays was accidental by Wilhelm K.
discovered.
Roentgen in Germany. Xrays were discovered while investigating the discharge ofelectricity through rarefied gases, the same phenomenon and equipment through which cathode rays were first noticed. At the time of researches going on cathode rays, Xrays were also being produced but Roentgen
3.1.1 Types of Xrays
According to thequality ofXrays, these can be divided intwo groups, these are
was the first person tonotice them when hewas investigating
(I) Soft Xrays: These are Xrays with low penetrating power.
the characteristics of cathode rays.
The wavelength range ofthese Xrays are from 10 Aand above hence their frequency is small and theyhave smaller energy.
As has been found when cathode rays (fast moving electrons) These are produced atcomparatively low potential difference
are braked bya material, their kinetic energy isconverted into electromagnetic radiations. These are known as Xrays. These are electromagnetic radiation with very short wavelengths extending from 0.01 Ato100 A. Since inthe visible region, the shortest wavelength are ofviolet rays which are visible tothe
and high pressure.
(II) HardXrays: These are Xrays having high penetrating power. The wavelength range ofthese Xrays are ofthe order of 1A so they have high frequencies and hence high energies. eye have wavelength ofabout 4000 A, Xrays are invisible to These are produced at comparatively low pressure and high eye.
potential difference.
Xrays are produced in Xray tube consist ofa glass or metal 3.2 Production Mechanism of Xrays envelope enclosing the assembly of a cathode, an anode (generally referred as target) spaced a certain distance apart
On thebasis ofproduction mechanism. Xrays are classified in
and connected to a source of extremely high tension (EHT)
two broad categories
supply. The cathode acts as a source ofelectrons and the anode
1.
Continuous XRays
as a source of Xrays. The field setup between the cathode
2.
Characteristic XRays
and the anode accelerates the electrons toenergies oforder lO'^ tolO^eV.
Nowwe'll discuss about the production mechanism of Xrays in detail.
The invisible Xrays are detected by observing their effects. Among other things Xrays produce a strong photochemical action which blackens photographic plates. They are also
3.2.1 Continuous Xrays
capable of ionizing gases and causing fluorescence in Continuous Xrays, as their name implies, have a continuous phosphors. For measurement purposes, these are used inmainly spectral distribution. They are produced when electrons for their applications oftheir photochemical and ionizing effects.
accelerated in a vacuum strike a target and lose kinetic energy
In ionization chambers, the intensityofXrays is determined by
in passing through the strong electric field ofthe target nuclei,
Roentgen discovered that the Xrays has following
3.2.2 Production of Continuous Xrays
characteristics.
Inthegeneral Xray production mechanism there isa discharge tube (generally called Coolidge tube) operated onhigh potential
measuring the saturation current due to the ionization of the andresulting in a continuous Xray spectrum. Asaccording to gas enclosed in the chamber, because the saturation current is the classical physics any accelerated charged particle emits electromagnetic radiation ofcontinuous spectral distribution. proportional to theintensity ofXrays.
(a) Xrays are generated whenever high energy cathode rays difference. The cathode used in the tube is the source ofcathode strike solid materials. Generally the greater the density of rays i.e. fast moving electrons, and theanode used in thetube impacted material the moreXrays areproduced. is thetargetmaterial whichis the source ofXrays. TheXrays
(b) Matter is more or less transparent to Xrays. Wood and
areproduced bythe bombardment ofanode material bycathode rays, so there will be a lot of heat generation in the anode
flesh areverytransparent bone andmetals lessso,which makes
material. The anode material should have high melting point as
the use of Xrays in medicine so useful.
(c) Xrays are unaffected (undeviated) byelectric and magnetic
duetohighenergetic electrons, a hightemperature isgenerated. For this purpose target is selected of high atomic mass. The coolidge tube assembly for the production ofXrays is shown
fields, hence these are uncharged.
infigure3.1.
'XRays
101
eV= ~ mv^
...(3.1)
2eV
...(3.2)
v = m
When this electron comes out fiom the atom ofanode material
the speed of this electron will be veryless as compared to its initial speed. Thus the difference of kinetic energy of this electron is emitted in the form of an Xrays photon fiom the
Xrays *
anode atom. Figure 3.1
The mechanism for the production of continuous Xrays can be explained on the basis of figure3.2. Figure showsan atom of the anode material of high atomic weight with its electron configuration. In the Coolidge tube an electron is projected towards the anodewith an accelerating voltageV. Sothe kinetic energy of the projectile electron will be eV. As shown in the figure when the projectile electrons enters into the extremely high electricfield ofthe nucleus ofthe atom of the anode material, it experiences a strong electric force towards the nucleus ofthe atom of the anode material, it experiencesa strong electricforce towards the nucleus of the atom and dueto this strong attraction the velocityof this electron when it emerges fi"om the atom will be highly reduced and negligibly compared with the initial velocity ofthe projectile electron. This electron in the influence of the highly positive nuclei experiences a very high acceleration
and according to the classical theoryeveryaccelerated charged particle emits the electromagnetic radiations, so this electron will also emit electromagneticradiations, these electromagnetic radiations are called Xrays. According to the law of conservation of energy, the energy of these electromagnetic radiation will be equal to the decreasein the kinetic energy of the projectile electron. This amount can be calculated. E = hv
Xray photon
Those electronswhich pass through the atom verycloseto the nucleus, will be more accelerated and the photon energy corresponding to these electrons will be more as compared to thoseelectrons whichpassthrough the atom at relatively large distance from nucleus. Themaximum energyofXphoton will be corresponding to that electron which looses almost all of its energy during passing through the atom. The photon correspondingto this electronwill have the shortestwavelength among all the photons radiated by other electrons. Ifthis shortest wavelength is then we have
AE= y1 ,
•)
=
he
12431 X
Xc = ^F7 ~ —77—A eV V
This is theminimumwavelength of Xrays emittedfrom an Xray tubewhich we call short wavecut off.Thus fiom equation(3.3) we can seethat the maximum energyor minimum wavelength of Xrays emitted depends only on the potential difference applied across the discharge tube.
Thus we can obtain Xrays inany range X^ toco byapplying an appropriate voltage across discharge tube which will fix X^ and other photons emitted firom tube will have wavelength more
thanX^ andranging uptooo. Thats why these Xrays arecalled continuous Xrays. The basic intensity per unit wavelength versus wavelength plot (wavelength spectrum) of continuous Xrays is shown in figure3.3.
Projectile Electron
Figure 3.2
If the initial velocity ofthe projectile electron is v, then as the velocity is gained due to the accelerating voltage V, we have
...(3.3)
Wavelength Figure 3.3
__
i
As we can see in the graph that the intensityof emittedXrays Ifcavity is created inKshell thenit maybefilled bytheelectron willbemaximum (maximumnumberofphotons) fora particular ofeither of L, M, N ... shells. If cavityis filledbyan electron of valueof wavelengthat a particular acceleratingvoltageacross Lshellthen the energyreleasedwill be equalto the difference thedischarge tube. Ata particular voltage theintensity ofXrays inenergies oftheKshell and Lshell. This iscalled AT^line and can be variedbychangingthe currentin the circuitbecausethe if the cavity is filled byan electron of Mshell then the line is
intensity of Xrays (number ofphotons) is proportional to the'
called Kpline. Similarly K^, Kg,... lines may also exist. This
number of electrons attacking the anode. The broad continuous spectrum beyond the peak intensity is referred as
series of lineris called Kseries of characteristic Xrays.
"Bremsstrahlung".
Ifcavity iscreated in Lshell thenaccording tothe transition of
3.2.3 Characteristic Xrays
this is called Lseries. There may also be Mseries as shown in
electrons from higher orbits there will be L^, L^,...lines, and
figure3.5(a). Figure3.5(b) shows thewavelength spectrum for TheseXraysare called characteristicXrays becausethey are the characteristicXrayswhen cavityis created in Kshell. characteristicofthe elementused as target anode. Characteristic Xrays has a line spectral distribution unlike to continuous Xrays. The wavelength spectrum of these Xrays is also a ° r \ Lseries continuous spectrum but this spectrum is crossed over by distinct spectrallines. The frequencies corresponding to these lines are the characteristic ofthe material of the target i.e. anode material.
3.2.4 Production of Characteristic Xrays
Production of these Xrays can be explained with help of the figure3.4. These Xrays are produced when the projectile
•
/ Mseries
electron towards the anode collides with an internal electron of
the atom of the target material, and then secondary emission will happen.Now this willcreate a cavityin the inner shellof the target material, and then secondaryemissionwill happen. Now this will create a cavity. There are so many ways in which the
(a)
cavitymay be filled. The cavity can be filled by an electron of higherorbitwhichmakethetransition fromitsorbitto the lower orbit in which cavity is created. When an electron make such a form of electromagnetic radiation, this energy is the characteristic line for the Xray spectrum.
Projectile Electron
Bremsstrahlung
(b) Figure 3.5
Projectile Electron
Thus the characteristic lines of the characteristic Xrays
depends only on the target material and not on the accelerating voltage. One more thing we can see here that the characteristic Orbital ^ Electron
Figure 3.4
Xrays are emitted only when the projectile electron make a collision with another bound electron ofthe atom ofthe target material. When cathode rays pass through the target then the
XRays
103
probabilityof collision of a projectileelectron to collide with the bound electron is very less. Majority of the projectile electron to collide with thebound electron is veryless. Majority
person is allowed to stand between a fluorescent screen and the Xray unit. The deep shadow of bones is formed on the
of the projectile electrons will pass through the anode without making the collisionand they produce continuous Xrays. So always both type ofXrays, continuous and characteristic will be emitted. Only a single type can not be emitted.
Instead of the fluorescent screen, a photograph known as radiograph can alsobe. taken. Xrays are alsousedto diagnose diseases in the lungs, kidneys, intestinesand other parts of the
fluorescent screen and the fracture of the bone can be detected.
body.
(2) Radiotherapy : Xrays are used to destroy malignant tumorsand to cure skindiseased. Longexposure ofXrays kills the germs in the body and hard Xrays are used to destroy
3.3 Moseley's Law
Moseley found that the wavelengths of characteristic Xrays depend in a well defined manner oh the atomic numbers of the tumors very deep inside the body. elementthat emitXrays (thetargetelement). Thisdependence, is known as Moseley law, may be written as (3) Industry: Xyrays are used to detect defect in radiovalves, tennis balls, rubber tyres and the presence of pearls in oysters. Jv =a{ZG) ...(3.4) They are also used to test the uniformity of the insulating materials and the quality of oil paintings. Where a is known as screening constant so the term (Z  a) here is effective atomic number for the electron which takes
part in the transition which results in characteristic Xrays and a is Moseley constant. It can be seen that for
line the electron make a transition
from Lshell to Kshell, and during this transition this electron move in the electric field ofthe nucleus and the one electron of
the Kshell, so the screening on the transition electron is only due to the remaining one electron of the Kshell so we can take the effectiveatomic number for this transition as (Z1) so the frequency emitted for the line can be given as
V^fl(Zl) The wavelength ofthe characteristic Xrays can be calculated by the relation
(4) Engineering : Xrays are used to detect cracks in structures and blow holes in metals. They are used to test the quality of weldings, moulds and metal coatings. They also help in detecting any crack in the body of the aeroplane and motor cars.
(5) Detection departments : Xrays are commonly used to detect the smuggling of precious metals at the custom posts and to detect the explosives and other contraband goods like opium in sealed parcels and in leather cases. They are also used in mints, where coins are made and every person has to pass before an Xray unit after the day's work is over. (6) Research : Xrays are used in research to study the structure of crystals, arrangements of atoms and molecules in matter and their behaviour on different materials.
{=R(Zof
1
1
For AT^line we use a = 1 and = 2 and wavelength of the AT^line can be given as
...(3.5)
= 1, thus the
# Illustrative Example 3.1
An Xrays tube operates at 20 kV.Find the maximum speed of the electrons striking the anticathode, given the charge of
electron = 1.6x 10^'^coulombandmassofelectron =9 x 10~^'kg. Solution
=>
v=fe(Zl)^
Thus for ^^linea =
, where 7? istheRydberg constant.
When an electron ofcharge e is accelerated through a potential difference V, it acquires energy eV. If m be the mass of the electron and v the maximum speed of electron, then
3.4 Applications of Xrays (1) Surgery: Xrays can pass through blood and not through bones. They are used to detect the fracture of bones, diseased organs and foreign bodies and growth in the human body. A
2eV m
XRays
1104
Dividing (3.6) and (3.7) we get
Substituting the given vales, we get
2x(1.6xl0"^^)x20,000 9x10
31
= 8.4 X10^m/sec.
>^1415A = 0.708A.
# Illustrative Example 3.2
(a) AnXray tubeproduces a continuous spectrum ofradiation with itsshortwavelength endat0.45 A.What isthemaximum energy ofa photon in the radiation ? (b) From your answer to (a), guesswhat order of acceleratingvoltage (for electrons) is required in such a tube ?
# Illustrative Example 3.4
If the short series limit ofthe Balmer series for hydrogen is 3644 A, findthe atomicnumber of the elementwhich giveXray
wavelengths down to lA. Identify the element. Solution .
Solution
Ifthe short series limit of the Balmer series is corresponding to transition « = coto w= 2 which is given by
(a) Short wavelength is given as
Maximum photo energy is given as he
^max ^max
Ti
'^min
12431 ^„.x=W=27624.44eV
The shortest wavelength corresponds to « = co to « = I. Therefore X is given as
^^27.624keV
J
(b) The minimum accelerating voltagefor electronsis 1
e
i.e.
l_
1
ofthe order of30 kV.
# Illustrative Example 3.3
The wavelength of the characteristics Xray
3644
line emitted
from zinc (Z= 30)is 1.415 A. Find the wavelength ofthe
= 911
Z1 =30.2
line
emitted from molybdenum (Z = 42).
Z=31.2:::3].
Solution
Thus the atomic number ofthe element is 31 which is gallium.
According to Moseley's law, the frequency for X series is given by
# Illustrative Example 3.5
A material whoseX absorption edge is 0.2 A is irradiated by Xrays ofwavelength 0.15 A.Find themaximum energy ofthe
v oc (Z 1)^
=>
f=c(Zl)^
photoelectrons that are emitted from the X shell. ...(3.6)
Where ^ is a constant. Let V be the wavelength of emitted from molybdenum, then
line
Solution The binding energy for X shell in eV is ho
...(3.7)
12431 0.2
eV=62.155KeV
!XRays 105
Theenergy oftheincident photon in eVis
# Illustrative Example 3.8
he _ 12431
X
0.15
82.873 KeV
The wavelength of
Xrays produced by a Xray tube is
Therefore, the maximum energy ofthe photoelectrons emitted
0.76 A. What isthe atomic number ofthe anode material ofthe
from the ^ shell is
tube ?
=£^^=82.873  62.155 KeV Solution
^
^max=20.718 KeV
Xrays are produced when an electron makes a transition # Illustrative Example 3.6
from w=2to «=1to fill avacancy in Kshell. The wavelength of Xraylines is givenby
Calculate the wavelength ofthe emitted characteristic Xray from atungsten (Z= 74) target when an electron drops from aM
1
={ziy •Ka
shell to a vacancy in the K shell.
j__j_ l2 22
1
^Ka
Solution
4
(Zl)2 =
Tungsten is a multielectron atom. Due tothe shielding ofthe nuclear charge bythe negative charge ofthe inner core electrons, each electron issubject toan effective nuclear charge which
3RX Ka
(Z\f =
3x(1.097xl0^)x(0.76xl0^®)
is different for different shells. = 1599.25 "
For an electron in the K shell (a = 1) thus effective nuclear
(Z1)2 ^1600
charge is given as
Z 1 = 40
2:eff=(Za);
Z=41.
# Illustrative Example 3.9
Here aselectron drops from Mshell {n =3)toX" shell (« = 1), the radiated emission wecall Xray andfrom Mosleys law the The Xabsorption edgeof an unknownelementis 0.171 A:
wavelength emitted ofXp Xray is given as l2
^Kp
32
10967800 X(741)2
(a) Identifythe element.
(b) Find the average wavelengths ofthe
and
lines.
(c) Ifa 100 eVelectron strike thetargetofthiselement, whatis the minimum wavelaigthoftheXrayemitted ?
''/tp
=0.192 A
Solution
# Illustrative Example 3.7
From Moseleys law, the wavelength oiK series ofXrays is given bytaking a = 1inmodified inrydberg's formula given as
Apotential difference of20kV is applied aXraytube. Findthe minimumwavelength of Xraysgenerated. Solution
Y=E.{Z 1)2 ^1—Lj for£:iineswhere,« =2,3,4,... (a) For ^Tabsorption edge, weput« = qo, inabove expression
The Xrays produced under an accelerating potential Vwill
gives
have varying wavelengths with the minimum due to the entire
energy ofthe accelerated electron being lostin a singlecollision with thetarget atoms. Here theshortest wavelength generated of Xrays can be given by X =
12431 V
z=
A
= 0.6215A
=>
1
(0.171xl0~'°)(1.097xl0^)
Z=74.
The element is Tungsten.
+ 1
XRaya
[106; Practice Exercise 3.1
(b) Forline: 1
(I)
= R{14\f
^^=0.228 A
[12.431 keV, 3 x lO'® Hz, 6.63 x 10'^"* kgm/s]
For Spline:
(II) What potential difference should be applied across an Xray tube toget Xray ofwavelength not less than 0.10 nm ?
=R{i4\y
Whatisthe maximum energy ofa photon ofthisXray injoule? [12.431 kV, 1.99 Xi0'5 J]
X.^=0.192A For
Find the energy, the frequency and the momentum ofan
Xrayphoton ofwavelength O.lOnm.
'Ka
line:
J—=R{14\f
1—r
^Ky
Xf^^=o.mA (c) The shortest wavelength corresponding toanelectron with
(III) Find the maximum potential difference which may be applied across an Xray tube with tungsten target without emitting any characteristic X orL Xray. Theenergy levels of thetungsten atom with an electron knocked outareas follows. Cell containingvacancy
K
L
M
Energy inkeV
69.5
11.3 2.3
[Less than 11.3 kV]
kinetic energy 100 eV is given by , _ ^
_ 12431 X
E
=>
100
^)
A free atom of ironemits
atom = 9.3 x 10"^° kg.
'X,=12431A
[3.9 X 10''" eV]
# Illustrative Example 3.10
(v)
Iron emits
Xray of energy 3.69 keV. Calculate the
timestaken by an iron
The
Xrays of energy 6.4keV.
Calculatethe recoilkineticenergyof the atom.Mass ofan iron
Xray emission line oftungsten occurs atX. = 0.21 A.
photon and a calcium
photon to
cross through a distance of 3 km.
What is the energydifference between K and L levels in this [10 gs by both]
atom?
(vl)
Solution
The wavelength of
Xray of tungsten is 21.3 pm. It
takes 11.3 keV to knock out an electron from the L shell of a
We know that
line is the radiation emitted when an electron
tungsten atom. What should be the minimum accelerating voltage across an Xray tube having tungsten target which
from Ishell{n = 2)makes a transition toAshell («= 1) tofill a allows productionof vacancy in it. Thus the released photon will have an energy equal totheenergy difference ofZ.shell and A^shell, which is [69.5 kV]
Xray ?
given by he
12431
''Ka
0.21
AE =
(vii) •eV
A£=59.195KeV
Theenergy ofa silver atom with a vacancy inK shell is
25.31 keV, inLshell is3.56keVandinMshellis 0.530keVhigher than the energyofthe atomwith novacancy. Findthe frequency
of
Ap and
Xrays ofsilver.
[5.25 X 10'® Hz, 5.98 x lo'® Hz, 7.32 x lo" Hz] Web Reference at www.Dhvsicsgalaxv.com
AgeGroup  High School Physics 1Age 1719 Years
Advance Illustrations Videos atwww.Dhvsicsgalaxv.com
SectionMODERN PHYSICS
Age Group Advance lilustrations
TopicXRays
Section  Modern Ph^^ics
Module Number  1 to 13
Topic  Atomicand Nuclear Physics Illustrations  54 Indepth Illustrations Videos
iXRaysI 107
Discussion Question Q31 In aCoolidge tube, electrons strike the target and stop Q316 Howare xraysproduced ?Explain the origin ofthe line inside it. Does the target get more and more negatively charged spectra and the continuous spectra. What limits the minimum as time passes? •
size ofXray wavelengths?
Q32 Can Xrays beused for photoelectric effect?
Q317 When asufficient number ofvisible light photons strike a piece ofphotographic film, the film becomes exposed. An Xrayphoton ismore energetic than avisible light photon. Yet,
Q33 Xray and visible light travel atthe same speed invacuum. Do theytravelat the samespeedin glass?
Q34 Characteristic Xrays may beused toidentify the element from which theyare coming. Can continuous Xrays be used
most photographic films arenotexposed bytheXray machines used at airport security checkpoints. Explain what these
observations imply about the number ofphotons emitted by the Xray machines.
for this purpose?
Q318 The drawing shows the Xray spectra produced by an Q35 Is it possible that in a Coolidge tube characteristic Xraysare emittedbut not
Q36 Can than
Xrays?
Xray ofone material have shorter wavelength
Xray ofanother?
Xray tube when thetube isoperated at two different potential differences. Explain why the characteristic lines occur at the
same wavelength inthetwo spectra, while thecutoffwavelength Xq shifts tothe right when a smaller voltage is used tooperate the tube.
Q37 Cana hydrogen atom emitcharacteristic Xray?
1 Higher voltage
Q38 Why isexposure to Xray injurious tohealth but exposure to visible lightis not, whenbothare electromagnetic waves?
Q39 When a Coolidge tube is operated for some time it
Bremsstrahlung
becomes hot. Where does the heat come from ? Wavelength
Q310 Can Xraysbe polarized ?
Q311 In terms of biological damage, ionization does more damage when you standin front of a very weak (low power) beam ofXrayradiation than in front ofa stronger beam ofred light. How does the photon concept explain this paradoxical
1
Lower voltage
X 
.
situation?
Q312 Whyshould a radiologist be extremely cautions about
Wavelength
Xray doses when treating pregnant women ? Figure 3.6
Q313 Does theconcept ofphoton energy shed any light (no pun intended)on the question of whyXrays are so muchmore penetrating than visible light ? Explain.
Q314 What isthebasic distinction between xray energy levels
Q319 In the production of Xrays, it is possible to create Bremsstrahlung Xrays without producing the characteristic
Xrays. Explain how thiscanbeaccomplished byadjusting the electric potential difference used to operate the Xray tube.
and ordinary energy levels ?
Q320 The short wavelength side of Xray spectra ends Q315 Can xraysbe emittedbyhydrogen?
abruptly ata cutoffwavelength X.^. Does this cutoffwavelength depend on the target material used in the Xray tube > Give your reasoning.
XRays
ri,08
ConceptualMCQs Single Option Correct 31 Xrays are produced when an element ofhigh atomic weight is bombarded by high energy:
(A) Protons (Q Neutrons
(B) Electrons P) Photons
38 With which characteristic of the target does the Moslems law relates the frequencyof Xrays ?
(A) Density (Q Atomic number
p) Atomic weight p) Interatomic space
32 Which of the following principle is involved in the generation of Xrays ?
39 Xrays areproduced in anXray tube operating at a given accelerating voltage. Thewavelength ofthe continuous Xrays
(A) Conversion ofkinetic energy into potential energy
has values from:
(B) Conversion of mass into energy (C) Conversion ofelectric energyintoradiant energy p) Conversion of electric energyinto em waves
(A) Otooo
33 In Xraytube when the accelerating voltage Vis halved, thedifference between thewavelengths ofK^ line andminimum wavelength of continuous Xray spectrum : (A) Remains constant (B) Becomesmore than two times (C) Becomeshalf
(B) \.
mm
to 00 where
mm
> 0
(Q Otok
310 The wavelength of Arays for lead isotopes Pb^^^ Pb^^^, Pb^*^ are Xj, are X3 respectively.'Then ; (A)
P) X,>X2>X3
•(C) Xjv P) X'>X;v'^ andhardXrays (Q Very had Xrays andlowfrequency yrays (D) Soft Xrays and yrays
37 Choose the correct statement:
(A) Energy of an atom with K shell electron knocked out is more than the energy of an atom with L shell electron knocked out.
P) Energy of an atom with L shell electron knocked out is more than energy ofan atom with K shell electron knocked out.
(Q Energy of
photon is the sum ofthe energies of an L
32 Let Xp and denote the wavelengths ofthe Xrays of and photon. the K^, Xp and lines in the characteristic Xrays for ametal: P) Total energy emitted in from ofX^ photons is more than (A)
(B)
the total energy emitted in from of Xp photon from an Xray setup.
(Q
X,
33 Two electrons starting from rest are accelerated byequal potential difference:
(A) Theywillhavesamekineticenergy (B) They will have same linear momentum
(C) Theywill havesamedeBroglie wave length (D) Theywill produce Xrays ofsame minimum wave length when they strike different targets.
3'8 Ina Coolidge tube experiment, theminimum wavelength of the continuous Xrayspectrum is equal to 66.3 pm, then (A) electrons accelerate through a potential difference ofabout 12.75kV in the Coolidgetube
P) electrons accelerate through a potential difference ofabout 18.75'kV in the Coolidge tube (Q deBroglie wavelength of the electrons reaching the anticathode is ofthe order of 10 mm
p) deBrgolie wavelength of the electrons reaching the anticathode is 0.01 A
34 When an electron moving at a high speed strikesa metal surface, which of the following are possible? 39 The potential difference applied to an Xray tube is
(A) The entire energy ohheelectron maybe converted into an
Xray photon.
Anyfraction oftheenergy oftheelectron maybeconverted into an Xray photon
(Q The entire energy of the electron may get converted to
increased. As a result, in the emitted radiation (A) the intensity increases
p) the minimum wavelength increases (C) the intensity decreases P) the minimum wavelength decreases
heat
p) The electron may undergo elastic collision with the metal
310 A beam of electrons striking a copper target produces
surface
35 Mark correctstatement(s):
Xrays.Its spectrum is as shown. Keeping'the voltage same if the copper targetis replaced with a different metal,the cutoff wavelength and characteristic lines of the newspectrum will
(A) circumference of orbit of an electron in Bohr's model is
change in comparision with old as:
equal to an integer multiple of deBroglie wavelength of the electron
P) Kinetic energy of electron increases with increase of principle quantum number
(Q when anXrayphoton isemitted, onlyenergy conservation law is satisfied
P) none of these
Figure 3.11
36 In anXray tube thevoltage applied is 20KV. The energy
(A) Cutoff wavelength may remain unchanged while
required to remove an electron from L shell is 19.9 KeV. In the
characteristic lines maybe different.
Xrays emittedbythe tube: (takehe = 12420eVA) (A) Minimumwavelength will be 62.1pm p) Energy of the charactersticXrays will be equalto or less
p) Both cutoffwavelength andcharacteristic lines mayremain unchanged.
than 19.9 KeV
different.
(Q LjjXraymaybe emitted P) L^Xraywillhaveenergy 19.9KeV
P) Cutoff wavelength will be different while characteristic lines may remain unchanged.
(C) Both cutoffwavelength and characteristic lines may be
XBays j
[114 311 Which of the following statements is/are correct for an Xray tube ?
(A) on increasing potential difference between filament and
(B) The minimumwavelength increases (C) The intensity remains unchanged . P) The minimumwavelength decreases
target, photon flux of Xraysincreases
(B) on increasing potential difference between filament and 315 Xray incident on a matoial: target, frequency of Xrays increases
(C) onincreasing filament current, cutoffwavelength increases (D) onincreasing filament current, intensity ofXrays increases
(A) P) (Q P)
Exerts a force on it Transfer energy to it Transfersmomentum to it Transfers impulse to it
312 The intensity of Xrays firom a coollidge tube is plotted
against wavelength Xas shown inthefigure3.12. Which ofthe
316 Regarding Xray spectrum, which of the following
following statements is/are correct:
statements is are correct:
(A) The characteristic Xray spectrum is emitted due to excitation of inner electrons of atom
P) Wavelength ofcharacteristic spectrum depend on potential difference across the tube
(Q Wavelength of continuous spectrum is dependent on the potential difference across tube P) None of these 317 Which ofthe following statements is/are false ? Figure 3.12
(A) On inaeea ngthez (atomic number) oftargetX,^ decreases (B) On increasing the accelerating voltage of tube  X^ increases
(Q On increasing the power of cathode, /q increases P) On increasing the power of cathode, decreases 313 For a given material, the energy and wavelength of characteristic Xray satisfy:
(A) EiKJ>EiK^)>E{K;) (C) X(KJ>XiK^)>HK^)
(B) E{MJ>EiLJ>E(K^) P) XiMJ>X{LJ>X(KJ
314 The potential difference applied to an Xray tube is increased. As a result, in the emitted radiation : (A) The intensity increases
(A) The energyofphotoelectrons emitted fiom a givenmetal by soft Xrays have less energy than those emitted by hard Xrays
p) To increase the intensity of Xrays, the filament current should be increased
'
(C) The characteristic Xrays have continuous range of wavelengths
P) Xrays werenamed so becauseof their mysterious nature at the time of discovery
318 Which of the following statements is/are true ?
(A) The wavelength of soft Xrays is less than that of hard Xrays.
P) Xrays are produced duringacceleration ofelectrons (C) The anticathodeis a metal of lowatomicweight P) The wavelength ofthe characteristic Xrays depends upon the nature ofthe metal of the target
•XRays 115
UnsolvedNumericalProblemsforPreparation ofNSEP, INPhO &IPhO Fordetailedpreparation oflNPhOandIPhO students can refer advancestudy material on www.physicsgalaxy.com 31 The Xray coming from a Coolidge tube has a cutoff 310 The electric current in an Xray tube (from the target to wavelength of80pm. Find thekinetic energy ofthe electrons the filament) operating at 40 kV is 10mA.Assume that on an hitting the target.
average, 1% ofthe total kinetic energy oftheelectrons hitting
Ans. [15.5 keV]
the target are converted into Xray. (a) What is the total power
32 Ifthe operating potential in an Xray tube is increased by
target every second ?
emitted as Xrays and (b) how much heat is produced in the
1% bywhat percentage does thecutoffwavelength decrease ? Ans. [Approximately 1%]
Ans. [(a) 4 W (b) 396 J]
33 The distance between the cathode (filament) and the target 311 The Xrays ofaluminium (Z 13) and zinc (Z= 30) in an Xray tube is 1.5 m. Ifthe cutoffwavelength is30 pm, find have wavelengths 887 pm and 146 pm respectively. Use theelectric field between thecathode and thetarget. Ans. [27.7 kV/m]
Moseleys law Vv =a(Z b) to find the wavelength ofthe Xray of iron (Z= 26).
Ans. [198 pm]
34 The shortwavelength limit shifts by 26 pm when the operating voltage in an Xray tube is increased to 1.5 times the
original value. Whatwasthe original approximate value ofthe operating voltage ? in KV. Ans. [15.9 kV]
35 The electron beam in acolour TV is accelerated through
312 Acertain element emits X^Xray ofenergy 3.69 keV. Use thedata from the previous problem to identify theelement. Ans. [Calcium]
^
certain elements are given below. Draw
32 kV and then strikes the screen. What is the wavelength of aMoseleytype plot ofVv versus Zfor
themostenergeticXrayphoton? Ans. [38.8 pm]
radiation
p Ca Mn Zn Br '
Energy (keV) 0.858 2.14 4.02 6.51
9.57 133
36 When 40 kV is applied across an Xray tube, Xray is 314 Use Moseley's law with = 1tofind the frequency ofthe obtained with a maximum frequency of9.7 x 10'^Hz. Calculate he value of Planck constant from these date.
X^ Xray ofLa (Z = 57) ifthe frequency ofthe X^ Xray of Cu (Z = 57) is known to be 1.88 x 10Hz.
Ans. [4.12 X 10"'^ eVs] Ans. [7.52 x 10'» Hz]
37 The Xp Xray ofargon has awavelength of0.36 nm. The
minimum energy needed toionize an argon atom is 16eV. Find 315 The
and Xp Xrays ofmolybdenum have wavelengths
the energy needed to knock out an electron from the X shell of
0.71 Aand 0.63 Arespectively. Find thewavelength ofZ^ Xray
an argon atom.
ofmolybdenum.
Ans. (3.47 keV]
Ans. [5.64 A]
38 An Xray tube operates at 40 kV. Suppose the electron 316 The wavelengths ofX„ and L Xrays ofamaterial are converts 70% of,ts energy,ntoaphoton at each collision. Find 21.3 pm and 141 pm respectively. Find the wavelength ofX„
the lowest three wavelength emitted from the tube. Neglect the xray ofthe material energy imparted to the atom with which the electron collides.
Ans. [44.3 pm, 148 pm, 493 pm]
P'"]
39 TheKaXrays ofmolybdenum haswavelength 71 pm. If
317 Heat at therate of200 W is produced in an Xray tube
the energy of a molybdenum atom with a X electron knocked out is 23.32 keV, what will bethe energyofthis atom when anL
operating at 20 kV. Find the current in the circuit. Assume that
only a small fraction of the kinetic energy of electrons is
electron is knock out ? In eZ.
converted into Xrays.
Ans. [5.82 keV]
Ans. [10 mA]
[116
^
,
318 Fill the blanks in each ofthe following statements:
XRays ]
by ahydrogen like element is 0.32 A. Calculatethe wavelenth of
Xp line emitted bythe same element. (a) In an Xray tube, electrons accelerated through apotential difference of 15000 V strike a copper target. Findthe speed of Ans. [0.27 A] the emittedXrays insidethe tube.
(b) Inthe Bohr model ofthe hydrogen atom, find the ratio of 321 Apotential difference of20 KV isapplied across an Xray thekinetic energy tothe total energy oftheelectron inaspecific quantum state. Ans. [(b)  1]
tube. Theminimum wave length ofXrays generated is Ans. [41]
322 CharacteristicXrays offrequency4.2 xlO^^Hzareemitted
319 Suppose a monochromatic Xray beam ofwavelength from a metal due to transition from L  to Kshell. Find the 10 pmissent through aYoung's double slitand theinterference atomic number ofthemetalusingMoseley's law. Take Rydberg pattern isobserved onaphotographic plate placed 40cmaway constanti?= 1.1 x lO'm"'. from the slit. What should be the separationbetween the slits so that the successivemaxima on the screen are separated by a
Ans. [42]
distance of 0.1 mm?
323 An Xraytube isoperated at20kV and the current through
Ans. [4 X 10"' m]
the tubeis 0.5 mA. Find the total energyfalling on the target per second as the kinetic energy of the electrons in J.
320 ThewavelengthofthecharacteristicXrayX^lineemitted
Ans. [lO]
4
Nuclear Physics andRadioactivity FEW WORDS FOR STUDENTS
In preceding chapters we have discussed about
structure ofan atom and different experiments involving properties of atomic structure. Now. we consider the composition and properties of the atomic nucleus. In this chapter we will discuss about the forces holding the nuclear
Alpha partides
Gold film
•Alpha partides 
Gold film
•
• —•
• . •
•
t
•
/ _ /
7
matter within the nucleus, the nucleus and the nuclear structure. We next examine the
"• e_ •
• —
»
~"Nudeus
(W
condition of nuclear stability and the natural
radioactive processes ofalpha, beta and gamma decay. Finally we considernuclear reactions and the energy involved in these reactions.
CHAPTER CONTENTS
4.1
CompositionandStructure ofThe Nucleus
4.5
Nuclear Reactions
4.6
Nuclear Fission
4.2
Nuclear Binding Energy
4.3
Radioactivity
4.7
Nuclear Fusion
4.4
Radioactive Series
4.8
Properties ofRadioactive Radiations
COVER APPLICATION
Flgur,e(a)
Flgure(b)
Nuclear reactors are today major source of power across the globe. Flgure(a) "shows a specific nuclear power plant in which power generation is done by nuclear reactors. Figure(b) shows the internal block diagram of a nuclear reactor.
Nuclear Physics and Radioactivity .
118
After Rutherford's discovery of the nucleus in 1911, scientists
(2) Isobars : Elements having same mass numbers and
slowly realized that the nucleus must be made of smaller different atomic numbers are called isobars. Thus isobars have constituent parts. Earlier it was proposed that the nuclei of same A, but different Z. atoms heavier then hydrogen consisted of hydrogen nuclei, called protons, and electrons. The protons would provide the (3) Isotones : The elements which have same number of mass andpositive charge andelectrons which were presumed neutrons are called isotones. Thus isotones have same {AZ). to be in the nucleus, would neutralize some of the charge. This theorycouldsatisfactorily account fora nucleusofatomicmass A and atomic number Zby assuming that the nucleus contained
(4) Isodiaphers: Those elements which havesame difference in neutrons and protons are called isodiaphers. Thus in
Aprotons andAZelectrons. The protonelectron model ofthe isodiaphers {A2Z) issameandthis value {A22) iscalled isotopic nucleusgavethe correctcharge and approximately the correct number of the element. mass for any nucleus. However this proton electron model of nucleushad manydifficulties. It wasnot ableto explainspin of 4.1.1 Size of a Nucleus particles emitted during decayprocesses from thenucleus. This
this model. For such reasons, Ruthorford suggested in 1920 that the nucleus should contain a neutral particle ofsame mass
The Ruthorford scattering experiment provided the first estimates of nuclear sizes. At that time a variety ofexperiments have been performed to calculate the nuclear dimensions. It was found that the volume of a nucleus is directly proportional
model does not conserve angular momentum in this
phenomenon. This factalonewasa strongreason fordiscarding
as that ofprotonbutnot ableto provepractically. Later in 1932
to the number of nucleons in it. Thus it can be said that the
James Chedwick demonstrate the existence of such a neutral
densityof nucleons is approximately same in the interiorsofall
particle in his experiments. This particle was given the name
nuclei.
neutron.
4
Now we'll discuss the ftindamental properties ofnucleus having protons and neutrons are the major constituents.
,
,
Ifanucleus is ofradius R, its volume is y %R., thus here R is directlyproportionalto the number of nucleonsor mass member A ofthe element. Hence we have
4.1 Composition and Structure, of The Nucleus Because protons and neutrons both appear in the nucleus, they are collectively called nucleons. According to today's accepted protonneutron model of the nucleus, the number of protons in a nucleus is called atomicnumber of the elementZ, and the number of protons plus the number of neutrons is called mass number ofthe element/4. Some timesy4 is also called
nucleon number. A shorthand notation is often used to specify Z and A along with the chemical symbol for the element. It is generallywritten as
R^ozA or
RozA^^^
or
R= R^ 1/3
Here
...(4.1)
is named fermi constant and its value is given as = 1.2x10"'^1.2 fin.
...(4.2)
As nuclei do not have sharp boundaries, the value of may also have slight deviations from its value given by equation(4.2).
2X or
4.1.2 Strong Nuclear Force and Stability of Nucleus Similarly in nuclear reactions the symbol used for a proton is
}p or [h . Aneutron is denoted by qii. In caseofelectron we
We've discussed that inside a nucleus in a very small volume
use _] e (It has 0 mass and its charge is  1). Someparticular
nucleons are bounded together. The positively charged protons inside the nucleus repel one another with a very strong electrostatic force. It is surprising that due to such a large repulsive force what keeps the nucleus from flying apart ? It is
names are given to different group of elements having some similarity in nuclear structure. These are (1) Isotopes : Elements those nuclei have same number of protonsbut differentnumber ofneutrons, are known as isotopes. Thus isotopes have same Z, but different X. Isotopes of same elements have same chemical properties because they have the same number and arrangement ofelectrons.
clear that there must be some kind of attractive force which
hold the nucleons together as many atoms contain stable nuclei. The gravitational force of attraction between nucleons is two weak to counteract the repulsive electric force, so we can say
that a different type of force must hold the nucleus together.
[Nuclear Physics and Radioactivity
119
This force is strong nuclear force and is one of the four fundamental forces ofnature;
We can see that in the graph shown in figure4.1 almost all points representing stable nuclei fall above the straight line N=Z, reflecting that the number of neutrons are greaterthen
The mostimportant feature of the strong nuclearforce is it is independent of electric charge. At a givenseparation between
number of protons.
nucleons, approximately some nuclear force ofattraction exist
In veryheavynuclei as numberofprotons are large,therecomes a point when balance of repulsive and attractive forces can not
between twoprotons, twoneutrons or between a proton and a neutron. The range of action of the strong nuclear force is extremely short. When two nucleons are very close at separation of the order of about 10"'^ m, the nuclear force of attraction between the two is very large and almost zero for large separations. For electric force we already know that the range of action ofelectricforce is verylarge.The electricforce
be achieved by an increased number of neutrons. For heavy nuclei, the size is also large and due to the limited range of strong nuclear force, extra neutrons in the nucleus cannot
balance the long range electric repulsion of extra protons. In naturethe stable nucleus withthe largestnumberofprotons is
Bismuth ^gj^Bi which contains 83 protons and 126 neutrons.
between two charges is very large at close separation and All nuclei above bismuth in nature are unstable due to decreases to zero gradually as separation increases to very unbalanced internal force and spontaneously break apart or large values.
The veryshort range ofstrongnuclear force playsan important role in stability of nucleus. For a nucleus to be stable, the electrostatic repulsion between protons must be balanced by the attraction between nucleons due to strong nuclear force.
rearrange the internal structure of nucleus. This phenomenon of spontaneous disintegration or rearrangement in internal" structure of nucleus is called radioactivity. This phenomenon wasfirst discovered byHenriBecquerel in ]896.In laterpart of this chapter we'll discuss radioactivity in detail.
4.2 Nuclear Binding Energy Inside a nucleus one protonsrepel all other protons. Since the electrostaticforcehas a longrange of action. Buta proton or a neutron due to strong nuclear force attracts only its nearest neighbours, thus in. a large sized nucleus to counter balance
the repulsive force more number of neutrons are required to maintain the stability ofnucleus.
We've discussed that in a stable nucleus, because of strong nuclear forceofattraction, the nucleonsare held tightlytogether in a small volume.As systemis stable we can relativelysay that
the totalpotential energyofsystem is negative and to separate all the nucleus from eachother someenergymustbe supplied tobreakthenucleus. Themorestable thenucleus is,thegreater is the energyneededto break it apart. This required energy, we call binding energy ofthe nucleus.
Binding energy
Nucleus
(smaller mass) Separated nucleons (greater mass) Figure 4.2
The origin ofbinding energycan beeasilyexplainedif we look into the masses of different elements in table4.1. In nuclear
Stable nuclei
physics, mass ofatoms and nuclei are often measured in atomic mass unit (amu). 20
40
60
Proton number Z
80
We can see clearly fiom the data given in table4.1 that the mass of each element atom is less then the sura of masses of its
Figure 4.1
constituent particles.
Nuclear Physics and Radloacttvity j
;i2o
Table 4.1
Element
z 0 1
2
3.
Neutron
Hydrogen
Helium
Lithium
Symbol
A
Atomic Mass, amu
z
1.008 665
14.
1
n
H
He
Li
1
1.007 825
2
2.014 102
3
3.016 050
3
3.016 029
4
4.002 603
6
6.018 891
6
6.015 123
7
7.016 004
8
8.022 487
7
7.016 930
15.
16.
*17. 4.
Beryllium
Be
•
6.
7.
8.
9.
Boron
Carbon
Nitrogen
Oxygen
Fluorine
B
C
N
F
.
10.
11.
12.
13.
Neon
Sodium
Magnesium
Aluminum
Ne
Na
Mg
A1
Sulfur
Chlorine
Atomic Mass, amu
Si
28
27.976 928
29
28.976 496
30
29.973 772
30
29.978 310
31
30.973 763
32
31.972 072
P
S
CI
33
32.971 459
34
33.967 868
35
34.969 032
36
35.967 079
35
34.968 853
36
35.968 307
37
.
36.965 903
Argon
Ar
36
35.967 546
10
10.012 938
37
36.966 776
11
11.009 305
38
37.962 732
12 ~
12.014 353
39
38.964 315
40
39.962 383
10
10.016 858
11
11.011 433
39
38.963 708
12
12.000 000
40
39.963 999
13
13.003 355
41
40.961 825
14
14.003 242
15
15.010 599
40
39.962 591
41
40.962 278
19.
20.
Potassium
Calcium
K
Ca
12
12.018 613
42
41.958 622
13
13.005 739
43
42.958 770
14
14.003 074
44
43.955 485
15
15.000 109
45
44.956 189
16
16.006 099
46
45.953 689
17
17.008 449
47
46.954 543
48
47.952 532
14
0
Phosphorus
A
10.013 535 18.
5.
Silicon
Symbol
9.012 182
9 10
Element
14.008 597
15
15.003 065
16
15.994 915
17
16.999 131
18
17.999 159
19
19.003 576
17
17.002 095
18
18.000 937
19
18.998 403
20
19.999 982
21
20.999 949
21.
Scandium
Sc
45
44.955 914
22.
Titanium
Ti
46
45.952 633
23.
Vanadium
V
47
46.951 765
48
47.947 947
49
48.947 871
50
49.944 786
48
47.952 257
50
49.947 161
51
50.943 962
18
18.005 710
48
47.954 033
19
19.001 880
50
49.946 046
20
19.992 439
52
51.940 510
21
20.993 845
53
52.940 651
22
21.991 384
54
53.938 882
23
22.994 466
24
23.993 613
54
53.940 360
55
54.938 046
22
21.994 435
24.
25.
26.
chromium
Manganese
Iron
Cr
Mn
Fe
54
53.939 612
56
55.934 939
57
56.935 396
23
22.989 770
24
23.990 963
23
22.994 127
58
57.933 278
24
23.985 045
59
58.934 878
58
57.935 755
59
58.933 198
60
59.933 820
25
24.985 839
26
25.982 595
27
26.981 541
27.
Cobalt
Co
Nuclear Physics and Radioactiwty z 28.
29.
30.
31.
32.
33.
34.
Element Nickel
Copper
Zinc
Symbol
A
Ni
58
57.935 347
96
60
59.930 789
97
96.906 018
6 1
60.931 059
98
97.905 405
100
99.907 473
Cu
Zn
Gallium
Germanium
Arsenic
Selenium
Ga
Ge
As
Se

35.
36.
Bromine
Krypton
Br
Kr
1
37.
38.
Rubidium
Strontium
121
Rb
Sr
Atomic Mass, amu
62
61.928 346
64
63.927 968
63
62.929 599
64
63.929 766
65
64.927 792
64
63.929 145
65
64.929 244
66
65.926 035
67
66.927 129
z
Element
Symbol
A
Atomic Mass, amu 95.904 675
43.
Technetium
Tc
99
98.906 252
44.
Ruthenium
Ru
96
95.907 596
68
67.924846
70
69.925 325
45.
Rhodium
Rh
69
^68.925'581
46.
Palladium
Pd
71
70.924 701
70 72
98
97.905 287
99
98.905 937
100
99.904 217
101
100.905 581
102
101.904 347
104
103.905 422
103
102.905 503
102
101.905 609
104
103.904 026
69.924 250
105
104.905 075
71.922 080
106
105.903 475
73
72.923 464
108
107.903 894
74
73.921 179
110
109.905 169
76
75.921 403
107
106.905 095
74
73.923 930
75
74.921 596
47.
Silver
Ag
•
48.
107.905 956
109
108.904 754
74
73.922 477
106
105.906 461
76
75.919 207
108
107.904 186
77
76.919 908
110
109.903 007
78
77.917 304
111
110.904 182
80
79.916 520
112
111.902 761
82
81.916 709
113
112.904 401
79
78.918 336
80
79.918 528
81
80.916 290
78
77.920 397
80
79.916 375
81
80.916 578
82
81.913 483
83
82.914 134
84
83.911 506
49.
50.
Cadmium
108
Indium
Tin
Cd
In
Sn
114
113.903 361
116
115.904 758
113
112.904 056
115
114.903 875
112
11 1.904 823
114
113.902 781
115
114.903 344
116
115.901 743
117
116.902 954
86
85.910 614
118
117.901 607
85
84.911 800
119
118.903 310
87
86.909 184
120
119.902 199
84
83.913 428
86
85.909 273
87
86.908 890
88
87.905 625
39.
Yttrium
Y
89
88.905 856
40.
Zirconium
Zr
90
89.904 708
52.
Antimony
Tellerium
Sb
Te
121.903 440
124
123.905 271
121
120.903 824
123
122.904 222
120
119.904 021
122
121.903 055
123
122.904 278
124
123.902 278
93.906 319
125
124.904 435
95.908 272
126
125.903 310
127
126.905 222
91
90.905 644
92
91.905 039
94 96
51.
122
41.
Niobium
Nb
93
42.
Molybdenum
Mo
92
91.906 809
94
93.905 809
95
94.905 838
92.906 378
53.
Iodine
1
128
127.904 464
130
129.906 229
127
126.904 477
131
130.906 119
Nuclear Physics and Radioactivity i
:12Z
z 54.
55.
56.
57.
58.
Element
Symbol
Xenon
Cesium
Barium
Lanthanum
Cerium
Xe
Cs Ba
La
Ce
A
Atomic Mass, amu
Holmium
Ho
165
68.
Erbium
Er
162
161.928 787
164
163.929 211
166
165.930 305
125.904 281
128
127.903 531
129
128.904 780
130
129.903 509
131
130.905 076
132
131.904 148
134
133.905 395
136
135.907 219
69.
Thulium
133
132.905 433
70.
Ytterbium
129.906 277
131.905 042
134
133.904 490
135
134.905 668
136
135.904 556
137
136.905 816
138
137.905 236
138
137.907 114
139
138.906 355
136
135.907 14
138
137.905 996
140
141.909 249
59.
Praseodymium
Pr
141
140.907 657
60.
Neodymium
Nd
142
141.907 731
143
142.909 823
144
143.910 096
62.
63.
Promethium Samarium
Europium
P m Sm
Eu
Gadolinium
Gd
167.932 383
170
169.935 476
Tm
169
168.934 225
Yb
168
167.933 908
170
169.934 774
171
170.936 338
172
, 171.936 393
173
172.938 222
176
173.938 873
71.
72.
Lutetium
Hafnium
Lu
Hf
74.
Tantalum
Tungsten
Ta
W
•
175.942 576
175
174.940 785
176
175.942 694
174
172.940 065
176
175.941 420
177
176.943 233
178
177.943 710
179
178.945 827
180
179.946 561
180
179.947 489
181
180.948 014
180
179.946 727 181.948 225
145
144.912 582
182
146
145.913 126
183
182.950 245
148
147.916 901
184
183.950 953
150
149.920 900
186
185.954 377
147
146.915 148

75.
Rhenium
Re
185
184.952 977
187
186.955 765
184
183.952 514
144
143.912 009
147
146.914 907
148
147.914 832
186
185.953 852
149
148.917 193
187
186.955 762
150
149.917 285
188
187.955 850
152
151.919 741
189
188.958 156
154
153.922 218
190
189.958 455
151
150.919 860
192
191.961 487
153 64.
166.932 061
168
174'
73.
61.
164.930 332
167
139.905 442
142
Atomic Mass, ai
67.
123.906 12
126
132
A
Element
124
130
Symbol
z
76.
77.
152
151.919 803
154
153.920 876
155
154.922 629
156
155.922 130
157
156.923 967
158
157.924 111
160
159.927 061
65.
Terbium
Tb
159
158.925 350
66.
Dysprosium
Dy =
156
155.924 287
158
157.924 412
160
159.925 203
161
160.926 939
162
161.926 805
163
162.928 737
164
163.929 183
78.
Osmium
Iridium
Platinum
Os
Ir
Pt
191
190.960 603
193
192.962 942
190
189.959 937
193
191.961 049
194
193.962 679
195
194.964 785
196
195.964 947
198
197.967 879
79.
Gold
Au
197
196.966 560
80.
Mercury
Hg
196
195.965 812
198
197.966 760
199
198.968 269
.
200
199.968 316
201
200.970 293
202
201.970 632
204
203.973 481
iNiiclear Physics and Radioactivity
Z
Element
Symbol
123
A
Atomic Mass, amu
Z
Element
Symbol
A
Atomic Mass, arau ~
81.
,82.
83."
Thallium
Lead
Bismuth
TI
,Pb
Bi
203
202.972 336
205
204.974 410
204
Polonium
Po
91.
Protactinium
Pa
233
92.
Uranium
U
232
232.037 168
233
233.039 629 234.040 947
203.973 037 205.974 455
234
207
206.975 885
235
235.043 925
208
207.976 641
238
238.050 786
210
209.984 178
214
213.999 764
237
237.048 169
239
239.052 932
209
93.
Neptunium
Np
——
233.040 244
206
212
84.
7
209.980 388 211.991 267
210
209.982 876
214
213.995 191
216
2i6.001 790
218
218.008 930
85.
Astatine
At
218
218.008 607
86.
Radon
Rn
220
220.001 401
222
222.017 401
87.
Francium
Fr
223
223.019 73
88.
Radium
Ra
226
226.025 406
89.
Actinium
Ac
227
227.027 751
90.
Thorium
Th
228
228.028 750
230
230.033 131
232
232.038 054
233
233.041 580
94.
Plutonium
Pu
239
239.052 158
240
240.053 809
95.
Americium
Am
243
243.061 374
96.
Curium
Cm
247.
247.070 349
97.
Berkelum
Bk
247
247.070 300
98.
Californium
Cf
251
251.079 581
99.
Einsteinium
Es
252
252.082 82
100.
Fermium
Fm
257
257.095 103
101.
Mendelevium
Md
258
258.098 57
Nobelium
No
259
259.100 941
103.
Lawrencium
Lr
260
260.105 36
104.
Rutherfordium
Rf
261
261.108 69
105.
Hahnium
Ha
262
262.113 84
• 102.
For examplewe discuss for 2He atom. Wecompare the mass of
Einstein mass energy relationship some amount of mass from
atom to that of its constituents. To calculate the masses of the
independent nucleons is converted into energy and released
components, we can either add the masses of two protons, two
when nucleons bounded with each other to form a stable
neutrons and two electrons or we can add the masses of two
nucleus. Thisis the energywhatholdthe nucleons together in
Hatoms and two neutrons. Using the values from table4.1, we
a nucleus, we call ''BindingEnergy' of nucleus. So when this energy is supplied to a nucleus, it splits into its constituent
have
particles.
2m^+2m^=2 (1.007825)+2 (1.008665) =4.0329804 •
...(4.3)
Nowwe can see that in table mass 2He atom is 4.0026034
For the above reaction given in equation(4.4), ifwe calculate the difference in masses of nucleons and that of nucleonsX, then it is given as
which is less then the value given in equation(4.3), the sum of masses ofconstituentparticles. Similarthing can be verifiedfor all the elements, also the reason for this can be explained by
This difference in masses of independent nucleons and mass
equation(4.4) which is a basic nuclear reaction for formation of
ofnucleus is called''Mass Defect' ofthe nuclear reaction. Using
Aw= Zmp+ (AZ)m
...(4.5)
a nuclei ^X. In this reaction we can see that when Zprotons mass defect Aw we can find the energy released in a nuclear and (AZ) neutrons fuse together to form a nucleus 2X, some
reaction, such as the binding energy ofabove nucleus X can be
amount of energy must be released as nucleus is more stable
given as
form ofnucleus
AEg^ =Awc^ ZWp+ {AZ)m
Energy
...(4.6)
...(4.4)
If we think fromwhere this energy come,we can simply say by
In thesimilarwaywecanfindthebindingenergyforanynucleus in nature for a known composition.
Nuclear Physics and Radioactivity :
ii24
4.2.1 Mass Energy Equivalence
AEv
AEy^
Ay
We'vediscussed in previous section that using massdefect of a nuclear reactionwe can find the energyreleasedin the nuclear
process. We know generally nuclear masses aregiven in atomic mass unit where
and these values aresuch that {BE^^
AE=0.092MeV
[9.435 MeV]
Because B is negative for this reaction, hence ^Be is unstable against decay to two alpha particles.
(vil) Find the binding energy of^gFe. Atomic mass of^^Fe is 55.934939 amu. Mass of a proton is 1.007825 amu and that of
;. Web'Referenceat www.physicsgalaxv.com
I Age Group  High School Physics [ Age 1719 Years
neutron = 1.008665 amu. [496.95 MeV]
: SectionMODERNPHYSICS
. Topic NuclearStructure &Radioactivity Module Number  1 to 16
4.3 Radioactivity As we've discussed that inside a nucleus electrostatic attraction
Practice Exercise 4.1
is counterbalanced by short range strong nuclear forces and nucleus becomes stable. Despite the forces are balanced, many
(!) In a thermonuclear reaction 1.00 x 10 ^kghydrogen is nuclides are unstable because of nuclear size or the limited range of nuclear forcesor dueto slight imbalancein small sized converted into 0.993 x 10"^ kghelium. (a)
Calculate the energy released in joule.
(b)
If the efficiency of the generator be 5%, calculate the
energy in kilowatt hours. [(a) 63.0 X lO'O J, (b) 8.75 kWh]
(ii)
nuclides, these nuclidesspontaneously disintegrateinto other nuclide. This phenomenon of spontaneous disintegration we callradioactivity. In further section ofthe chapterwe'll discuss the aspects of radioactivity by which unstable nuclide disintegrate to achieve stability.
Find the binding energy and the binding energy per 4.3.1 MeasurementofRadioactivity
nucleon ofthe nucleus of gO. Given, atomic mass of^gO (w) =
15.994915 amu, mass ofproton (wp—1.007825 amu, mass ofa Radioactivity ofan element is measured in terms of"activity" neutron (m„)=1.008665amuandlamu=931.5MeV. Theactivityofasampleofanyradioactivenuelideistherateat [127.62 MeV, 7.976 MeV]
which the nuclei ofits constituent atoms disintegrate. If are
Nuclear Physics arvd Radioactiyityi
1128
thenumberofnucleipresentin a radioactive sampleat an instant Thus due to randomness, approximately the amount of disintegrations perunitvolume persecond (called disintegration then activity of this sample is given as
density) remains constant in thewhole volume ofthesubstance. dN
...(4.8) 43.3 Radioactive Decay Law
Here
dN
is negative as with time always no. of elements
This lawrelatesthe activityof a substance withthe number of decreases due to disintegration, due to negative sign, is active or undecayed atoms present in a group of radioactive always takenpositive. The activity of a substance is measured atoms at an instant oftime. This law is stated as in termsof"dps" or disintegrations per second. The SIunit of "TTie activity ofa radioactive element at anyinstantisdirectly activity is named after Bequeral, defined as 1bequeral= 1 Bq=\ disintegration/sec
proportional to the number ofundecayed active atoms {parent atoms) present at that instant.''^
Generallyactivities of radioactive samples in nature are very highthat's whybequeral isa verysmall unitfornormal practice, Letus consider that at r= 0,thereareA'q parent atoms arethere moreoften MBq or GBq are used.Forthe sametraditionalunits in a substance and after a time ?, A^atoms are left undecayed. curie (C/) and rutherford{Ru) are alsoused. These are defined Thisimplies that in theduration from t = Q\.ot = t,N^Natoms are decayed to their daughter element. If in further time from as tt\.ot = t + dt,dN more atoms will decay then at time r = r,we can say that the activity of the element is given as
1 Ci = 3.7 X lO^'dis/sec and
1 Ru = 10^dis/sec
dN
Roughly curie was originally defined as activity of 1 gm of
^Ra. Similarly itwas observed that 1kg ofordinarypotassium
...(4.9)
A =
Now according to Radioactive Decay Law, we have
hasan activity of about 1 raCi (10~^ Ci) because in ordinary potassium smallproportion ofradioisotope
is alsopresent.
dN_ dt
xN
43.2 Fundamental Laws of Radioactivity ^ .
dN A =
Onthebasisofexperiments performed byRuthorford and Soddy
dt
= XN
...(4.10)
Here Xis the proportionality constant, we call decay constant for the decayprocess. The value of decay constant differs for we summarize as fundamental laws ofradioactivity. These are different elements. From equation(4.10) we can see that ifX, is high the elementwillhave high value ofactivityand ifXisless, (1) Radioactivity is purely a nuclear process, it is not the activity will be relativity less. Thus we can say that the concerned in any manner with the extranuclear part of atom. decayconstantfor a radioactive elementgives.a relativecriteria (2) As radioactivity is a nuclear process, it is independent of its stability ifthe value of Xfor an element is more, it is more fromany chemicalpropertyof the element.As we'vediscussed active or relatively less stableand iffor an elementXis less, it is that radioactive property of an element is only the process more stable.From equation4.10, we can also write
some conclusions were made for behaviour of radioactive
elementsand the properties ofRadioactivity, these conclusions
concerned with nucleus of the element. It does not effect the
electronic configuration of the element. If this element takes partin a chemical reaction theproduct formed willalsohavethe radioactive property in the same fraction by which the radioactive atom is present in the molecule of product.
(3) Radioactivity is a randomprocess, its studyis onlypossible by laws of probability mathematically. In a group of several radioactive atoms, which one will disintegrate first is just a matter of chance.
dN dt X=
Thus decayconstant of a processcan be given as "activityper atom" (as given in equation(4.11)). This shows that for a given radioactive element the activity per atom always remains
constant where as we've already discussed that with time the overall activity of a substance decreases with time as number of parent elements continuously decreases with time.
(4) As radioactivity is a random process, the disintegration densitythroughout the volumeofa radioactive element remains Now from equation(4.10), we can write constant. If an elementX decaysto a daughter nuclide Ythen in a given volume of element, all portions of volume will have same ratio of number of atoms of Y to that ofX
...(4.11)
N
dr=~^
...(4.12)
•Nuclear Physics and Radioactivity
"12^1
Herenegative sign shows that ~ , the rate at which the active Insome nuclear power plants amajor problem isdisposal ofthe
elements are disintegrating is negative or number of active elements are decreasing with time. Now we have from equation(4.12)
radioactive wastes since some ofthenuclide present in waste have long halflives.
For a radioactive element in a sample if at / = 0, No nuclei are present ofactive parent element and during observation after dN
//
, ,
^dt
t= ^ are leftthen thisduration 7canbetaken as halflifeof
Now we integrate this expression within time limits from /=0, to this element. From radioactive decay equation wehave t = t,WQ have
N=NQe \r
N
...(4.15)
t
IfI
Here a.t tT,N= ~
Xdt
thus wehave inabove equation
N,
[inivE =Xt
In I^I=XT
\nN\nN^ ='kt In I
\n(2)=XT
Xt
^ In (2) _ Q.693 N=N^e^'
...(4.16)
...(4.13)
Here equation(4.13) gives thenumber ofactive parent atoms N 4.3.5 Alternate formofDecay Equationin terms ofHalfLife present attimet inthemixture. Thisequation, wecallradioactive Time decay equation.
If one radioactive element X decays to a daughter nucleus Y with a decay constant Xthen for theprocess nuclear reaction, is
From equation(4.13) we can have
written as
U[=XN^e'^' A c =A cO
.X/
X
^
Y
...(4.14)
Ifinitially nuclei ofelement Xare present then after time t, Here A^^ = XN^is the initial activity of substance at r = 0. number of nuclei present in the sample are given by decay equation(4.14) is another form ofradioactive decay equation. equation given as This equation can be used to find activity of a radioactive
N=N^e^
substance at any time instant.
...(4.17)
If werearrange the equation,we have 4.3.4 HalfLife Time
In ^ =XT
In previous section we've discussed that during decay of a radioactive sample, the amount of radionuclide fall off
exponentially with time. Every radioactive sample has a
^0
We know that half life of substance is defined as
y,^In(2)
characteristichalflife. Halflifetime is defined as the time duration
in which halfof the total number ofnuclei will decay or left undecayed. Say for example at any instant we look into the quantity of a sample of radioactive element, it is observed that
X or
we can write
In (2) ...(4.19)
after every3 hour the number of undecayed parent element reduces to halfthus accordingly the half life of the nuclide is 3 hour. Some half lives are only a millionth of a second for highly active elements and some less active elements have half life in billions ofyears.
...(4.18)
Now from equation(4.18) we have In
jV_
ln(2)
Nr.
T
Nuclear Physics and Radioactivityj
:i3o
time t is dN. The sun in the number ofthe average is really an
ln^l=ln(2r"
integration ofthe quantity t ^iVbetween
and 0particles; the
denominator is the sumofthe particles, or the integration ofdN over all the particles:
Taking antilog on both sides, we get idN
— =(2r"^
A^o
^^
...(4.22)
Mean or average life = —
N=N^{2)IIT
I
...(4.20)
dN
^^0
Equation(4.20) is an alternate form ofdecay equation useful pQj. fornumerical applications.
integration in the numerator, we use the activity ofthe
substance to find dN, given as
4.3.6 Mean Life Time at
Fora given radioactive substance, some nuclei disintegrate in dN= XN dt thebeginning andsome willdisintegrate aftera longtime. Thus wecansaythat somenuclei haveveryshortlifeandsomehave We make the substitution for dN in the numerator ofequationverylargelifetimefordisintegration. Fora radioactive element (4.22).For the averagelife,we then have mean life is defined as
1 tXN dt
Mean life time
T
>"
=
Mean or average life =
Sum of lives of all nuclei in a sample Total number of nuclides in sample
r= m X
...(4.23)
_ N,
From radioactive decay equation we have
NNff'^'
...(4.21)
The above numerical value is comes out to be reciprocal of
Thus equation(4.23) becomes CO
decay constant oftheelement. This cannotbedirectly calculated
dt
but we can calculate the value ofmean life time for a radioactive
element using basic probabilitylaws as explained in the next
Mean life 
N,
...(4.24)
section.
This integration is done by parts. The result comes out as 4.3.7 Calculation of Mean Life Time For a Radioactive Element
The halflife ofa sampleofradioactive nucleiis thetime forhalf ofthe sample to disintegrate. The average or mean life of the
sample isdifferent. Some oftheatoms in thesample exist much longer than others before decaying. To determine themean life, consider an analogy. Imagine a collection of 10people withthe following death statistics; 3 die at age 60 y, 2 at age 70 y, 4 at 75 y, and 1 at a venerable 90 y. The average age is found by multiplying thenumber dying ateachage, summing theresults, and dividing the sum by the total sample size : Average age =
3(60y) + 2(70y) + 4(75y) + l(90y)
...(4.25)
Mean life =
Thus the numerical value ofmean life is the reciprocal ofthe
disintegration constant for the respective radioactive element. # Illustrative Example 4.7
The halflife ofradon is 3.8 days.Afterhow many dayswill only one twentieth of radon sample be left over ? Solution
10
•ny
The average lifetimeofan initial sample ofradioactive atoms is found in a similar way. LetA^be the number ofatoms that still
We know that
Here
=0.693/7
7=3.8 day
exit at time t. Between t and t + dt, we lose a few ofthese hearty
atoms: dN of them decay. Thus the number of atoms that live a
%=
0.693 3.8
= 0.182 per day
jNuciear Physics and Radioactivity
131
Ifinitially atr= 0, the number ofatoms present be N^, then the number of atoms N leftaftera time t is given by
log
'21100
N
20 =>
50
T=
Nn
100
log2
99
^ r=50x
e^=20
logio(2) Iog,o(100)Iog2(99)
Taking log on both sides, we get 21.9956
2.303logio 20 2.303 logio 20 0.182
0.301
. T=50
X/ = ln{20)2.3031ogio20
T= 3420 sec = 57 min.
16.45 days.
# Illustrative Example 4.8
One gm of a radioactive material having a half life period of 2 years is kept in store for a duration of4 years. Calculate how much of the material remains unchanged ?
§ Illustrative Example 4.10
1 g of a radioactive substance disintegrates at the rate of
3.7 XlO"^ disintegrations per second. The atomic mass ofthe substance is 226. Calculate its mean life. Solution
Given that activity of substance is Solution
^^=3.7xi0"0dps According to radioactive decay law we have
The number of atoms in 1 gm ofsubstance are
N=Nq2"'^ N=
Here we have T=2 yrs. and t = 4 yrs.
Thus we have
N=Nq2
=>
1x6.023x10^^ 226
N=2.66 X10^' atoms
If X is the decay constant of the substance, we know that
KJ 1 iV=^ = ^gm
Activity
AXN
Thus after 4 years 0.25 gm of the material will be left.
^ N
a Illustrative.Example 4.9
1gm of radioactive substance takes 50 sec to lose 1 centigram. Find its halflife period.
ao
.
=>
X=
=>
3.7x10'
^ s'
2.66x10^'
A.= 1.39 X 1011 s"^
Thus mean life of the radioactive substance is Solution X
Given that after 50 second the amount remaining is 0.99 gm as out of 1 gm, 1 centigram is lost. Now fiom radioactive decay T
equation, we have
=
I ^11
1.39x10"
N=N^2*''^ 0.99 = (1)250/r
r_ = 7.194xl0i°s
Nuclear Physics and Radioactivity!
!:132
No. ofatoms in 2.5 mg ofsubstance are
Illustrative Sample 4.11 There is a stream ofneutrons with akineticenergyofO.0327 eV. If the half life of neutron is 700 seconds, what fraction of
neutrons will decay before they travel a distance of 10 km.
(mass oftheneutrons= 1.6758 x 10"^'kg).
2.5x10"^ x6.023x102^
^ =>
230
Ar=6.54 XlO^^atonK
We know that activity is given as • A=XN
Solution
or
decay constant is
Given that kinetic energy ofneutrons is
1= ^ ^ N
=0.0327 X(1.6 X10"*^ J
8.4 X= .2
6.54x10
2x0.0327x(1.6xl0"^^) 1.675x10
^
v^= 625xl0'^
=>
v = 2500m/s
18
X=1.28xl0'®s^
27
Thus half life of substance is
ln(2) T=
Time to travel a distance of 10 km
0.693 T=
lO'^m t =
2500m/s
,18
1.28x10" = 4s
7=5.41 xio'^
After 4 second number of neutrons left can be given as
# Illustrative Example 4.13
N=Nq2"
When n= yt no. ofhalflives. Here nJL
= 2
4
1
1/175 _
= 0.996
after two days ?
A^=0.996N„
Thus fraction of neutrons decayed is Nq~N
In an experiment on two radioactive isotopes of an element (whichdonot decayintoeachother), theirmass ratio at a given instant was found to be 3. The rapidly decaying isotopes has larger mass and an activityof 1.0 p. curie initially.The halflives of the two isotopes are known to be 12 hours and 16 hours. What would be the activity of each isotope and their mass ratio
Solution
0.004Aro
f=
= 0.004
Let the two isotopes are A and B such as their halflives are
r^ = 12hr. a Illustrative Example 4.12 and
An experiment is done to determine that halflife of a radioactive substance that emits one beta particle of each decay process.
7B = 16hr. m.
Initiallytheir mass ratio is given — =3
[As
Measurements show that an average of 8.4 beta particles are
emitted each second by 2.5 milligram of the substance. The atomic weight of the substance is 230. Find the half life of the substance.
Given that initially activity ofisotope^ is 1 pCi
^^=lpa Activity ofisotope A after 2 days will be given as
Solution
Given that activity of the substance is A = 8.4 dps
A =^ 16'
> W5]
tNuclear. Physics and Radioactivrty
133
Aftert = 5 hrs, the activity presentin the solution is
iLlC/
=>
 0.0625
=>
^=3.7xl0''x2'^3dps
For the second isotope B it atomic mass assumed to be same
the initialactivity canbe given as
Given that in 1cm^ ofblood sample is
dps. Iftotal value of
blood is Vthen total activity is lii{2)
ln(2)
4)^x12
16
3xln
296
3.7x10"
60
,1/3
v=
=>
"^OB
After two days activity of second isotope willbe
3.7xl0"x60 296x2'^^
cm
K=5952cm^ = 5.9521tr.
# Illustrative Example 4.15 In an ore containinguranium,the ratio of
A)B
Aqj 32
A
32
to ^®^Pb nucleiis
3.Calculate the age oftheore, assuming that allthelead present in the ore is the final stableproduct of
Takethehalflifeof
to be4.5 X10^ years. Solution •
^5 = 0.03125 pCi
No. of atoms of isotopes A andB left after two days are
Presentlythe ratio of
to ^®^Pb nuclei is 3
N.
N;,=NA2r
N
Ni=N,(2r' ^'a
1
=3 Pb
Ifwe consider no. ofuranium atoms are
3 Then no. of lead atoms
# Illustrative Example 4.14
Given thatall lead atoms arethedecayproduct ofuranium thus
Asmall quantityofsolution containing Na^radionuclide (half in the beginning at ?= 0, no. of uranium atoms can be taken as life15hours) ofactivity 1.0microcurie isinjected intotheblood ofa person.A sample ofthe bloodof volume 1 cm^ taken after 5 hours shows an activity of 296 disintegrations per minute.
N,u=AN,
If areoforeis t thenfi^om radioactive decay equation, we have
Determine thetotalvolume ofblood in thebody oftheperson. Assume that the radioactive solution mixes uniformly in the
N^=N,^{2r'''
blood of the person (1 curie = 3.7 x 10'® disintegration per second). w/4.5 X10®_ 4
(2)
Solution
Giventhat initial activityis
/ = 4.5xl09
^ = ipa
=>
^ = 3.7xl0'®xi0^dps
=>
^ = 3.7x lO'^dps
yrs
logio(2) ^
_ 4.5x10^x0.125 0.301
r= 1.868 xloVs.
Nuclear Physics an^ Radioactiyityj
UZ4 _ _ ;
After time t no. of atoms of equation as
# Illustrative Example 4.16
can be calculated by decay
N,,=N,,,{2r"
In the chemical analysis of a rock, the mass ratio of two radioactive isotopes isfound tobe 100:1. Themean lives ofthe
two isotopes are 4 X10^ years and 2x 10^ years respectively. If
N 14
N,014
it is assumed that, at thetime offormation ofthe rock, the atoms
An '012
An •"'012
ofthe two isotopes were in equal proportion, estimate the age
hit
(2)
[As Aq,2 remain unchanged]
of the rock. The ratio of the atomic weights of the two isotopes
1.1 X10'2 = 1.2 X10^2(2)''^
is 1.02:1.
1.2
Solution
Given that the presentmass ratio of the twoisotopes is m,
^=100
Iogio(2)
m2
The ratio of no. of atoms can be given as N,
m =— X^ m2
N2
= lOQx
0.0377 t =
1
ICQ
1.02
1.02
0.301
xSVaOyr
x5730yr
= 719.1 yr.
From radioactive decay equation, we have
and
N^ =N^^e'"''
..(4.26)
j Web Reference at www.phvsicsgalaxv.com
^2=^20^"'^''
...(4.27)
Age Group  High School Physics  Age 1719 Years j SectionMODERN PHYSICS
Dividing equation(4.26)by equation(4.27) 1
^ Topic  Nuclear Structure &Radioactivity ; Module Number16 to 30
1
1
In
A
Practice Exercise 4.2
N,
t =
1
1
2.3031og
(!) The half lifeperiodof radium is 1590 years. Afterhow manyyearswill one gram of the pure element. (a) be reduced to one centigram, (b) loseone centigram.
100 1.02 yrs
t =
I
1
[10564.78 years, 23.06 years]
.2x10^ 4x10^ J = 1.833 xlO'Vs
(ii) The disintegration rate ofa certainradioactive sampleat any instant is 4750 disintegrations per minute. Five minutes latertheratebecomes 2700disintegrations per minute. Calculate
# Illustrative Example 4.17
A bone suspected to have originated during the period of Ashokathe Great, was found in Bihar. Accelerator techniques
the halflife ofthe sample. [6.135 minutes]
14.
gave its
C
— ratio as 1.1 x 10
Is the bone old enough to
C
have belongedto that period ? (Takeinitial ratio of
= 1.2X10"^^ andhalflifeof"'C = 5730 years). Solution
Given that initial ratio of'''C to iVn '014 N,012
is
= 1.2x10"'^
with
(iii)
Calculate the activity of one gm sample of jgSrwhose
halflife period is 28.8 years. [5.106 X 10'^ dps]
(iv) The halflife ofa cobaltradioisotope is 5.3 years.What strength will a millicurie source of the isotope have after a period ofone year. [0.87 mCi]
[j^clear.Physics and Radioactiy^^^ (v)
Aisample contains 10"^ kgeach oftwo substances Aand
that the heavy nuclides having mass number A = An, where n is
5 with half lives.4 seconds and 8 seconds respectively. Their an integer, can decay into one another in descendingorder of atomic masses are in the ratio of 1:2. Find the amounts ofA and
mass number constituting a radioactive series. Thus there can
B after an interval of 16 seconds.
befour radioactive series having mass number specified by4w,
[6.25 X 10"^ kg, 2.5 X 103
,
...
(vQ At a given instant there are 25% undecayed radioactive nuclei in a sample. After 10secondsthe number of undecayed nuclei reduces to 12.5%. Calculate (a) meanlife ofthe nuclei, and (b) the time in which the number ofundecayed nuclei will further reduce to 6.25% ofthe reduced number.
An + 1, 4« + 2 and An + 3. Table4.2 lists the four radioactive
series. Amongall series,ttiehalflifeofNeptuniumis soshortas compared to the age of solar system that the members ofthis
series are not found on Earth. This series is produced in laboratory by bombarding other heavy nuclei with neutrons. From table we can see that all three natural radioactive series
terminate with the end product lead (Pb)and the artificial series terminatesin Bismiith(Bi)
[(a) 14.43 s, (b) 10 s] Table 4.2
(vil) 1gram ofcesium137 ClJCs) decays by pemission with a halflife of 30 years. What is (a) the resulting isotope? (b) the numberofatoms leftaftet 5years? (c)Ifinitialactivity ofsample is 1 mCi then what is the activity ofcesium after 5 years ? [(a) "^Ba, (b) 3.913 x IQ^', (c) 0.89 mCi.]
Mass Numbers
Series
Parent
An
Thorium
•^^Th
An + \
Neptunium
AiJ+2 An + 3
Uranium Actinium
^jNp
Half  life, y Stable End Product ' 1.39x10'" 2.25x10"
238,, 92U
.4.51x10"
235,,
7.07x10*
92''
^Pb
(viil) The^riormal activity ofa livingmatter containing carbon is foundto^be 15 decays per minute per gram of carbon. An archaeologicalspecimen gives6 decaysper minute per gram of carbon. If the halflife of carbon is 5730 years, estimate the approximate age ofthe specimen. [7575.40 years]
Figure4.4showsthe uranium series{A = An '+2 series)which starts from the parent nucleus and terminates in All the intermediateelementsofthe seriesare shownin figure. We can see here that some of the radioactive nuclei have
alternate decay modes, decaying by either alpha emission or
betaemission like^'®Po, ^'''Bi, ^'^i etc.
(ix) • A radioactive isotopeA'hasa halflifeof3 second. Initially a given sample ofthis isotope contains 8000 atoms. Calculate (a) its decay constant, (b) the time when 1000 atoms of the
isotopeXremain in the sample, and (c) thenumberofdecayper second in he sampleat/ = fj. 1(a) 0.23i'j', (b) 9 s; (c) 2318'] 
4.4 Radioactive Series
In the discovery of radioactivity we know it was formed that many of radioactive elements are relatively heavy. It was seen that sometime when one nucleus decays into another, the resulting daughter nucleus is also unstable or radioactive and it subsequently decays into another daughter nucleus until a stable element appears. This series ofall daughter nuclides we call radioactive series and the end product of the series is a stable isbtope. There are four radioactive series discovered and majority of radioactive elements can be considered to be members of one
a decay \
ofthese radioactive seri^. Out ofthese four, three are naturally found radioactive series and one is artificial laboratory made series. The reason that there are exactly four series follows from
p decay
thefact that when aradioactive elenient decays byadecay, the mass number ofnucleus reduces by 4. Due to this we can say
Figure 4.4
[136^^ _
Nuclear Physics and Radioactivity
4.4.1 Radioactive Equilibrium
Similarlyfora radioactive elementwithdecayconstantXwhich decays by both a and pdecay is given that the probability for
In a radioactive series ifwe talk about an intermediate element,
an a emission, isP, and that for pemission is
it is produced due to the decay of its previous element and it decaysto the next element of the series. In the equation{4.28)
constant of the element can be split for individual decaymodes, like in this casethe decayconstantsfor a and P decays separately
shown an intermediate element J decays to K with a decay
constant X.j &K decyas to L withdecay constant J
K
N,
N.
^
L
then thedecay
can be given as
...(4.28) and
Ifat an instant, iVj nuclei ofJ arepresent, itsdisintegration rate 4.4.3 Accumulation ofa Radioactive Element inRadioactive can be given as
1
Series ...(4.29)
As J decays to K, the above relation in equation(4.29) also gives the formation rate of nuclei of K. If at this instant nuclei ofK are present its decay rate can be given as
In a radioactive series, we've discussed that each element decays into its daughter nuclei until a stable element appears. Consider a radioactive series shown below A,
A.
^3
...(4.30)
To analyze mathematically the above series, we assume initially If at some instant the production rate and decay rate of the element becomes equal then the amount ofi^ appears to be a constant as the number of nuclei ofATproduced per second are equal to the number of nuclei oiK disintegrating per second.
at /• = 0,Nq atoms ofparent elements, arepresent which decays to the element ^2 ^ decay constant Xj, thus after time t, number ofundecayed nuclei ofA, present at a timeinstantt can be given be decay law as
This situation for the intermediate element AT is called radioactive
equilibrium. We can also state that this equilibrium is a dynamic equilibrium in which the amount ofK element appears to be a constant along with the process ofits continuous formation by decaying element /and its continuous disintegration to element L. Thus for an element condition of radioactive equilibrium is
N^=N^e^i'
:..(4.32)
Dueto disintegration ofA^, nuclei ofA^ are formed and these startdecaying with a decay constant X2 to another element Ay Let at an instant t, undecayed nucleiofA^ are presentthen the decay rate ofA^ at this instant can be given as
Rate offormation = Rate ofdisintegration Decay rate of Here fore element K to be in radioactive equilibrium, we have
=
...(4.31)
A~= X.N^
...(4.33)
Dueto disintegration of.4j, A2 isproduced thustheproduction rate ofnuclei ofA^ will bethedecay rateofnuclei of.4pthus production rate of.^2 instant canbe given as
4.4.2 Simultaneous Decay Modes of a Radioactive Element
Production rate of A^ = X, We know that due to radioactive disintegration a radio nuclide transformed into its daughter nucleus. Depending on the nuclear structure and its unstability a parent nucleus may undergo either a or p emission. Some times a parent nucleus may undergo both types of emission with the probabilities of adecay or Pdecay. The amount ofdaughter nuclide produced by a and pdecay will be in the probability ratio of a and p decays.
...(4.34)
Now in a further time dt^ if 0^2 nuclei of element A2 are accumulated thenthe accumulation rateofnuclei ofelementy42 can be given as
dN2 dt
= X,W,X2W2
dN.
...(4.35)
Now we analyze the decay ofsuch elements which disintegrates with two or more decay modes simultaneously. If an element decays to different daughter nuclei with different decay
solving gives the numberofnucleiofelement yfjas a function
constants A,p
oftime t. On solving equation(4.35) we get
... for each decay modethen the effective
Equation(4.35) is a simple linear differential equation which on
decay constant of the parent nuclei can be given as
XiWo N
2
= ——
(X1X2)
(gVe^^i')
...(4.36)
iNuclear Physics and Radioactivity' 137
Here we can see that in the begging as
= 0, due to #IllustrativeExample4.19
disintegration of^p^2 being formed and as the amount of^, is decreased and that of increases and that of
is increased, decay rate of
The mean lives ofa radioactive substance are 1620 years and
decreases. After a time when both 405 years for aemission and pemission respectively. Find out
decay rate becomes equal, the element ^^2 will be said to be in the time during which three fourth ofasample will decay ifitis
radioactive equilibrium and later the amount of A^ start decaying both by aemission and Pemission simultaneously.
decreasing with time. Thus it is the state of radioactive
equilibrium when the yield ofthe radionuclide A^ is maximum. Lets take some examples to understand the above phenomenon better.
# Illustrative Example 4.18
Solution
The decay constants for a and p emissions are 1/1620 and 1/405 per yearrespectively.
In this case effective decay constant for both decays
A radionuclide with half life 14.3 days is produced in a reactor at a constant rate ^ = 2.7 x 10^ per second. How soon
simultaneously is
after the beginning ofproduction ofradionuclide will its activity be equal to^ = 1x 10^ disintegration/sec
1
Solution
Let t be time in which the given sample decays three fourth. Therefore, the fraction of sample undecayed in timet is 1/4.
In the reactor just after production of radio nuclide, it starts
Hence
decaying. The accumulation rate of the radio nuclide can be given as
• N=N^J4 Nowfiom decayequation
.
A.
dN
qXN JV
=^dt
I
r_^
r
J qXN
J
dt
t=
ln(4)
= 1.386x324=449yr
# Illustrative Example 4.20
qXN = t
Lead ^°^Pb is found in acertain uranium ore due to disintegration
_ ^ X/ qXN=qe
of uranium. What is the age of uranium ore if it now contains
0.8 gm of^°^Pb for each gram of
given the halflife of
uranium=4.5 x 10^ years.
When activity
Solution
1 x 10^ dps then 14.3
206 grams of Pb is producedby238 grams of
, (2.7
' In(2) "'"U.V 14.3
,
Hence 0.8 gramsof Pbwillproduced by
{21
log,„(2) '"Sioll? 14 3
^ X0.8 =0.92427 grams of^^^U Now fiom decay equation we use w = Wg
/'= 9.55 days.
1 = 1.92427
^
Nuclear Physics and
•138,
Here
1
•e T= halflifeperiod andN=number ofmolecules,
Whe
= 1 + 0.92427 = 1.92427
m,
Further,
= ^(0693 ;/4.5 X10^)
mass of the sample x Avogadro's number
1.92427
N=
0.693?
4.5x10^
mass number
= In (1.92427)
Let tie total mass ofuranium mixture be M.
= 2.303 log,0 (1.92427)
Thep, mass of^92^,
4.5x100.693
... —rx^M 0006 ,, Mi=
[2.303 log,0 (1.92427)]
100
mass of^9U,
=4.25 xlOV
0.71
^2=
# Illustrative Example 4.21
Find the halflifeofuranium, given that3.32 x 10 ^gmofradium isfound pergmofuranium inold minerals. The atomic weight
and
mass of^92^, 99.284 ,.
ofuranium and radium are 238 and 226 and halflife ofradium is
1600years(AvogadroNumberis6.023 x 10^^/gmatom). Again,
0.006 ,. 6.03x10^^ Arj= ^^Mx234
100
Solution
Invery old minerals, theamount ofanelement isconstant this implies that theelement exist in radioactive equilibrium thus
0.693
and
N,
Nr
^
and
Mx
6.03x10
23
238
100
R
235
100
7.1x10^ 99.284
myAR
234
0.693 ..0.71M,,6.03x10^^ X 'X •
R.=
'u
and
0.693 ..99.284M..6.03x10"
Ri—
^ 4.5x10^"
100
Thus activities are in the ratio
^ ^
1x226
0.006
X 1600yr
0.71
234x(2.5xl0^) ' 235x(7.1xl0^)
3.32x10""^ x238
0.01026:0.000426
T,,=4.7xl0V
99.284
# Illustrative Example 4.22
Asample ofuranium is a mixture ofthree isotopes
• 238x(4.5xl0^)
'^IpJ
and^92U present in the ratio of0.006%, 0.71% and 99.284%
respectively. The halflives ofthese isotopes are 2.5 x 10^ years, 7.1 X10^ years and 4.5 x 10^ years respectively. Calculate the contribution to activity (in %) of each isotope in this sample.
: 0.00927
Totil activity = 0.019956 %A
ctivities,1^^x100=51.41% 0.000426
Solution
Activity
23
0.71., 6.03x10^^
XuNy and
6.03x10
'"2.5x10^'' 100 „
here we can use
0.006M
0.019956 0.00927 0.019956
X 100=2.13%
X 100=46.45%
238
[Nuclear Physics and Radioactivity
139
# Illustrative Example 4,23
iptN)E^
A radionuclide with half life T is produced in a reactor at a constantratep nuclei per second. Duringeach decay, energy
=ptE.
is released. If production of radionuclide is started at r= 0,
pEqT In
(1e^O
# Illustrative Example 4.24
calculate
Nuclei of a radioactive element A are being produced at a
(a) rate of release of energy as a function of time
constant rate a. The element has a decay constant X. At time
(b) total energy released upto time t
t = 0, there are
nuclei ofthe element.
(a) Calculate the number N ofnuclei ofA at time t
Solution
(b) If a = 2Nq X, calculate the number ofnuclei of^ after one As productionrate of radionuclideisp, the accumulation rate of these nuclide in the reactor can be given as
halflifeofA, and also the limiting valueofiVas t > oo Solution
[Where
X= (a)
=
dN
pXN = di
Rate of decay =  X N, and rate of formation = a, thus accumulation rate ofelement is dN dt
'plN'
dN = t
P
aXN
.
=dt
Integrating this expression, we get
p~XN
=>
= aXN
XN=p(le'^')
In
.(4.37)
Thus after time / the activity ofradionuclide in the reactor is
aXN = Xt
aXNo aXN
aTJ^Q
A^ =XN=p{le'^')
aXN^(aXNQ)e^'
Given that during each decay an energy Eq is released, thus rate ofenergy release or power ofreactor P at time / is
N= Y [a(aXNQ)e
P^A.^E. In. =pEf^(l~e^)' [Where A.^]
(b) \fa = 2Nr,X
N=j[2N^X{2NqXN^ X)e^']
Upto time /, number of undecayed nuclei can be given by equation(4.37) as
At the time ofhalflife,
In a time /, total number of nuclei produced are pt. Thus upto time /, number ofnuclei decayed are
N^=ptN Thus total energy released upto time / is
Et=N^>^E^
T=(p.693/X) So,
N=N^,[2e^^^']=^
Limiting value ofN (as /
oo)
N=N,[2e^]=2N,.
Nuclear Physics and Radioactiyity;
!140
(v)
; Wjeb Reference afwww.phvsicsgalaxv.com
^'Co decays to ^^Fe by P+emission. The resulting ^^Fe
is in its excited state and comes to the ground state by emitting
? Age: Group  High School Physics  Age 1719 Years I;Section, IvlODERN PHYSICS ' Topic i^udear Structure&Radioactivity
7rays. Thehalf life of p^ decay is 270 days and that of the
yemission is 10"® sec. Asample of^^Co gives 5x10^gamma raysper second. Howmuchtimewillelapsebefore the emission
rate ofgamma rays drops to2.5 x 10^ per second ?
Mbdule Number  31 to 37
[270 days]
Practice Exercise 4.3
(vO
Aradionuclide Ajgoes through thetransformation chain
(1) A radio nuclide with decay constant >,[ transforms A, >Aj >Aj (stable) with respective decay constants X, into a radionuclide with decay constant 7^2' Assuming that and Xj Assuming that at the initial moment thepreparation at the initial moment the preparation contained only the radio contained only the radionuclide A^ equal in quantity to Ajq nuclide Aj. Consider initially therewere nuclei ofA[ were nuclei, find the equation describing accumulation of the stable there at/=0, find:
isotope Aj.
(a) The equation describing accumulation of radio nuclide d2wifiitime. (b)
The time interval after whichthe activityof radio nuclide
^2 reaches itsmaximum value. 4.5 Nuclear Reactions [(a) m
=
j Kn
A) ~Ai
(11) Bi can disintegrate either by emitting an aparticle or by emitting a pparticle.
(a) Write the two equations showing the products of the decays. (b) The probabilities ofdisintegration by a and Pdecays are
intheratio 7/13. Theoverall halflife of^^^Bi isone hour. If 1g ofpure ^'^Bi istaken at 12.00 noon, what will bethe composition ofthis sample at 1 p.m. the same day ? T1 + a, ^11 Bi ^ 212 Bi ^ 2Upo + e + V, (h) 0.50
When two nuclei came close to each other, nuclear reaction can occur that results in a new nuclei being formed. Generally it is
very difficult to.bring nuclei very close as they are positively charged and repulsion between them keeps them beyond the range where they can interact unless they are moving very fast toward each other to decrease the distance of their closest
approach to start the nuclear reaction. In Sun nuclear reactions are very frequent as the temperature is millions ofkelvin, ftie nuclei have sufficient speed to bring then very close to each other and the energy released in reaction maintains the temperature there.
g  Bi, 0.175 gTI, 0.325 g  Po]
(iii) A^^®Bi radionuclide decays via the chain ^'®Bi ^
Most of the nuclear reactions occur in two stages or steps. In
first step an incident particle strikes a target nucleus and the two combine to form a new nucleus called compound nucleus,
206p^ (stable), where the decay constants are
which have atomic number and mass number equal to the sum
X, = 1.6 X10~^s""\X.2=5.8 X10"®s"'. Calculatea&p activities
of target nucleus and the striking particle. The compound nucleus is generallyfohned in excitedstate becauseof the initial kinetic energy of the striking particle. It is observed .that the
^"^0 A.2
of the ^'^Bi preparation of mass 1.00 mg a month after its manufacture.
W =A^oX, exp. (X0 = 0.72 Xio"part/s,^„=
compound nuclei have very short life time oftheorder of10"'^ seconds. In second step these compound nuclei may decay in one ca: more different ways dq5ending on their excitation energy.
. (iv) A human body excretes (removes by waste discharge, sweating etc.) certain materials by a law similar to radioactivity. If technetium is injected in some form in a human body, the
body excretes halfthe amount in 24 hours. A patient is given an
Lets discuss an example when an aparticle incident on a N nucleus with some kinetic energy. The corresponding nuclear reaction is given in equation(4.38).
injection containing ^^Tc. This isotope is radioactive with a + ^H
>'lF
>^0 + 1H
...(4.38)
halflife of 6 hours. The activity from the body just after the injection is 6 pCi. How much time will elapse before the activity falls to 3 pCi ?
Here fluorine nucleus
[4.8 hours]
some excess energy which automatically lead it to eject a particle
F, formed in an excited state. This has
IKTudear Physics and Radioactivity
141 j
like in this case, a proton (h ). As the excited nucleus lasts There aretwo major categories ofnuclear reactions, we'll discuss only for a short time, it iscommonly omitted from theequation
in detail. These are
ofnuclear reaction thus above reaction can also be written as
(i) Nuclear fission
N
^ 'bO + Ih
+
(iQ Nuclear fusion
...(4.39)
lets discuss these in detail.
Nuclear reaction such asthose written inabove have a general form
4.6 Nuclear Fission A + a
...(4.40)
> B + b
Figure4.5 shows the variation ofBinding Energy per nucleon
Here uppercaseletters represent the nuclei and the lowercase letter represent the particles. The shorthand form of above
with mass number of differentelement. Wehave discussedthat
the middle weight elements are more stablethen the heavy weight elements. Thus if we can break a large nucleus into smaller nuclei energy must be released. Ifthe released energy is ...(4.41)
reaction is written as
A(a,b)B
morethen that required,to split the nucleusthen the difference
For the example we've taken above equation(4.39) can be energy can be converted into useful forms. written as
',^N(a,p) 'lo
Thephenomenon ofsplittinga heaviernuclei intotwo or more lesser weightfragments is knownas nuclear fission.
...(4.42)
4.5.1 QValue of Nuclear Reaction
For example in the figure say
0value ofa nuclear reaction is defined as the difference in the rest mass energies of particle and nuclei before reaction and
energy required is
Binding energyper nucleon
those ofproducts formed after reaction. Forexample inanuclear
is a heavy element with then to break it into nucleons
C/j = 150£^
reaction.
...(4.45)
Ifnuclei splits into two fragments Yand Zaccording toreaction A + B
givenin equation(4.46)
...(4.43)
C + D
The 2value of above reaction is defined as
150
'X
Q=im^ +m^m^mj)(?
56 t:
75. 33
...(4.46)
and Z is
a reaction g is a negative quantity then for reaction to take place energy must be supplied to the system.
26
^
90
...(4.44) Then the amount of energy released in formation of nuclei Y
Ifin areaction energy isreleased, ^ isa positive quantity. Ifin
31B
> 60y ^
U2 =60Ej.+ 90E^
...(4.47)
209R,
4tt j/w59
19
/Ll2c
^
I
50
92"
 3^Li
• 3 eo
.5 CO 1
Nucleon number A
Figure 4.5
Nuclear Physics and Radioactivity]]
142
Where Eyand
are the binding energy per nucleon ofnuclei
YandZ respectively. Thus theenergy released in thefission of nucleusXper. fission Af" can be given as
i,n +
^ flu* ^ "{Ba +
+3l,n ...(4.50)
The above reaction is one out of many possible reactions like
AE=t/217i
!in + fU
=60£y+90.E^150£^
...(4.48)
'SXe +
+2!,n ...(4.51)
But it is observedthat the maximum yield of the reaction is for
The above energy A£' or gvalue of this reaction can also be equation(4.50). Some reaction may produce as many as 5 neutrons but the average number of neutrons produced per calculatedbymass defectAw of this reaction as fission is 2.5.
AE = Anic^
Fission processes have the possibility of some practical uses because wehavepositive ^values ofthereaction. Thepositive The fission fiagments formed byfission ofa heavy nuclei areof 2value occurs because the heavy element break up into two unequal size becausethe heavy nuclei have a greater neutron fragments that aremore tightly bound andtherefore have less toproton ratio as compared to lightnuclei, thusthefragments total mass energy. ...(4.49)
= {nixmy
will have more neutrons. Hence to achieve stability generally
out of twofission fragments, one is a middle weightfragment Aheavy nucleus undergo fission when it hasenough excitation and other is a relatively heavy element because of excess energy so that the compound nucleus oscillate violently. For neutrons in the heavy nucleus. To reduce this excess, two or example (which isabout 99.3% innatural uranium) undergo three neutrons are emittedbythe fragments as soonas theyare fission reaction onlybyfast neutronswhosekinetic energyare formed and subsequent beta decays followed by yemission more then 1 MeV. Whereas few nuclei like are able to split adjust their nip ratio to stable values. into twonucleibyjust absorbing neutron. For suchcases very slowneutrons are required to start fission reaction as in fast neutrons the contact time ofneutron with the nuclei is very less
Fission
to start fission with such nuclei.
fragment
Another important aspect of nuclearfission is the amount of large magnitude ofenergy released. Forexample during fission
Neutron
A. f I'X" Neutron
of each uranium nuclei about200 MeV energyis released. In all fission reactions majorfraction of the energyreleasedis in the formofkinetic energyofthe fragments. Roughlyabout80 to 85
percent ofthe energy released is carried bythe fragments of fission as their kineticenergy. About23 percentis in the form ofkinetic energy of neutrons. About 25 percentis in the form ofemitted yrayphotons duringfission andthe remaining1015
Fission
fragment
Q Neutron
percent ofthe total energy is the form of subsequent betaand gamma decays ofthe fission fragments.
Figure 4.6
4.6.1 Fissionof Uranium Isotopes and Chain Reaction In 1939 Otto Hahn and his colleagues found that a uranium nucleus, after absorbing a neutron, splits into two fragments each with a mass smaller then the original nucleus. Figure shows
a fission reaction in which a uranium "I Unucleus issplit into
barium 'jgBaand krypton ^gKrnuclei. The reaction starts when a ^9! U nucleus isabsorbs aslow moving neutron (kinetic energy  0.04 eV or less)wecall thermal neutron,transforms
into a compound nucleus, ^9! U. This compound nucleus have
avery short life time and disintegrates quickly into 'sgBa and 3g Kr and three neutrons. The reaction can bewritten as
In each fission we've discussed that an average of about 25 neutrons are released. Without these neutrons practically the utilization of the fission energy becomes difficult. If we place
the fissioning nuclei quantity in a properamount it is possible to obtain a self sustained fission reaction. The neutrons from the fission can cause other fission. Each fission releasing more
neutrons and energy as shown in figure4.7. Such a process is called a chain reaction. If such a process occurs in an uncontrolled fashion, it results in a very big explosion. If the some of neutrons from each fission are stqjped by some external mechanism to limit the uncontrolled chain reaction then the
kinetic energy(now limited) offragments can be extracted and utilized to do useful work. This process is carried out in nuclear fueled power plants.
[Nuclear Physics and Radioactiwty
143
The neutrons lose their kinetic energy by collisions with moderator and theyslowdown initial theirkineticenergy is of the other of thermal neutrons to start further fission events.
Detailed explanation ofnuclearreactoris not given here as in earlier classes you have studied about a nuclear reactor in detail. » Lost
Weadvise you to refer the same once again.
neutron
Lost  neutron
4.6.2 Liquid Drop Model Lost neutron
Nuclear fission of uranium can be understood on the basis of
liquid drop model ofnucleus. When a neutron strikes the nuclei, it is captured by the nucleus and it is transformed into
neutron
the excited which due to excess energy oscillates in a varietyofways. Figure4.8(a) shows the stages of oscillations ofthe asa liquid drop. The drop inturn becomes a spheroid,
a sphere, an oblate spheroid, a sphere and a prolate spheroid First
Second
Third
Fourth
generation
generation
generation
generation
neutrons
neutrons
neutrons
again and so on.
Figure 4.7
The major problemin abovecasesof energyproductionis, the products ofthe fission process are extremely radioactive. These are in majorityyemitters with large lives, whichis a wizardfor
Time
(a)
health.
Most ofthe nuclear reactors use uranium as their nuclear fuel.
As we've discussed that in natural uranium only 0.7% is rest is In chain reaction plays an important role. Thusbefore usinginnuclear reactor naturaluranium isprocessed and enriched to about 3.5% of the isotope
Some students may think why^^^U do not participate inachain
(b) Figure 4.8
Oscillating stages of compound nucleus
Theinertia ofthe moving liquid molecules causes the drop to overshoot sphericity and go to the opposite extreme of
reaction. The answer is simple.As we've discussedthat to start
distortion. It is observed the nuclei exhibit surface tension and
fission in
we need very fast neutrons having kinetic
these can vibrate like a liquid drop when in excited state. For
energies more than about 1 MeV. Other wise when struck by a neutron, it captures the neutron to become and
small excitation it is possible to oscillate like this as distortion in nucleus is recovered back by its surface tension. But when
according to equation(4.50) it decays to byemitting two excitation is large and distortion in nucleus is large,once the p particlesand immediately fissions whenstruckbya slow excitation energyof nucleus is given off in the form ofyray moving thermal neutron. The reactions photon, the surface tension due to shortrange strongnuclear forces is not able to bring back the nucleus into its spheical 4. 238tt 0n + 92 U '9IU + Y ...(4.52a) shape and the nucleus splits into two parts as shown in 239
92 239
U 
93 Np
239
93
Np + _oe + Y
> 'I'Pu + 0' _'ne +
...(4.52b) ...(4.52c)
In a nuclear reactor, the probability that ofa neutron will cause
to fission is more for a slow moving neutrons. Thus there is less probability that the fast neutrons produced in fission will cause other fission to occur and chain reaction will start. To
slow down these neutrons (to initiate more fission events) in a nuclear reactor moderator is used which may be ordinary water
carbon or heavywater(D2O).
figure4.8(b). 4.7 Nuclear Fusion
If we again look at the figure4.9 which shows variation of binding energy per nucleon ofelements with mass number, we
can seethat for heavyweightelementsaveragebinding energy per nucleon is about 7.8 MeV and for middle weight stable elements it is about 8.6 MeV. Thus we can say that the average energy released per nucleon by fission is the difference
between these two values about 0.9 MeV per nucleon.
Nuclear Physics
il44
Radioactivity]
For another example consider a general fusion reaction, given as
2
+ 3^
»
Here 2 nuclei ofan element Fission
+ AE(fusionenergy)
and3 of^B fuse together toform
a single middle weight nuclei
Fusion
...(4.54)
If we know the values of
binding energy pernucleon for theelements A, B and CasE^,
and E^then the amount ofreleased energy can be given as 100
A£' = 38£^10£^18£^
150
Nucleon number A
...(4.55)
Equation(4.55) is an alternative wayto calculatefusion energy ifthe bindingenergypernucleon forall the elements of reaction
Figure 4.9
are known. If masses of individual nuclei are known AE can
Now if we take a look on the left portion of this curves, the
steepness is relativelyhigh or wecan saythat the differencein bindingenergypernucleon between verylightnuclei andmiddle weight nuclei are much more compared to heavyand middle weight elements. Thus if two verylowweight nuclei are fused together to produce a middle weight nuclei having greater binding energy, the energy released per nucleon will be very high compared to energyreleased pernucleon in a fission event. This process is callednuclearfusion. For example lookat the fusion reaction given in equation(4.53) •H
+ tn
> ;He
+
In a fusion reaction, reaction will start only when two nuclei are
brought sufficientlycloseto each other so that the short range nuclear force can pull then together. We know at large
separations coulomb force dominates and due to strong electrostatic repulsionbetween positivenuclei, it is very difficult to bring nuclei so close that strong nuclear force dominates and start the fusion reaction. This distance at which fusion
starts is called coulomb potential energy barrier and it is ofthe 0"
...(4.53)
Here two deuterium isotope fuse together and releases
approximately 4 MeV energywhichwe can easilycalculateby using mass defect of the reaction. This energy is about the same energy per nucleon released in a fission event. But the most important thing is most of the fusion reaction products such as 2He here are stable and non radioactive. Proton
also be calculated by mass defect of the reaction.
Neutron
2
Tritium jH
Deutenum jH nucleus
nucleus
(contains two neutrons)
(contains two neutrons)
order of 10"^'' m. At such close distance nuclei will approach each other only when they have large kinetic energies which is possibleonly at high temperatures of the order of hundreds or thousands ofmillion kelvin.
Such reactions which takes places only at such high temperatures are called thermonuclear reactions. Under such conditions all the atoms are completely ionized in gaseous form. Such a high temperature gas of positive and negative charged particles is called plasma. The main problem in nuclear reactions is to confine plasma for a long enough time so that collisions among the ions can lead to fusion. There are three methods of confining the plasma named, magnetic confinement, inertial confinement andZpinch which is a slight modificationof inertial confinement. Detailed analysis of these methods is beyond the scope of this book.
Fusion
# Illustrative Example 4.25 On disintegration of one atom of the amount of energy obtained is 200 MeV.The power obtained in a reactor is 1000 kW. How many atoms are disintegrated per second in the reactor ? What is the decay in mass per hour ? Solution
Neutronqh
Helium oHe
nucleus
Figure 4.10
Power produced in the reactor is =>
E = 1000kW=1000x lO^W
[Nuclear Physics and Radioactivity =>
liii
P=10^J/s
The number ofatoms of^^^U consumed in 1000 hoursare
A^, = 10'^x{l000x3600) =36xl0^
10^
P =
1.6x10"'^
eV/s
Now 1 gmatoms of
mass of^^^U consumed in 1000 hours ofoperation are
= 6.25xl0'SMeV/s
Asineach disintegration 200 MeV energy isreleased, number
w
of atoms disintegratedper secondare N=
6.25x10'^
has 6 x 10^^ atoms. Therefore, the
,
=>
=3.125 xloS'
36x10^^ ^
6x10^^
x235 = 1410gm
®
m = 1.41 kg
# Illustrative Example 4.27
Energyreleased per second is 10^ J
• The fission type of war head of some guided missiles is
Energy released per hour is
estimated to be equivalent to 30000 tons of TNT. If
E= 10^x60x601
3.5 X10® joules ofenergy are released by on tone ofexploding TNT how may fissions occur and how much
Thus mass decay per hour can be given as
would be
consumed intheexplosion ofwarhead ?Anenergy of200 MeV Aw  —Y (Einstem s massenergyformula)
is released by fission of one atom
c
Solution /Sm =
10''x60x60J
(3xl0%/s)^
Energy released by war head =3000 tons ofTNT (3.5x10®) (30000) = 10.5 Xl0'2j
Aw^4x 10"^kg Aw =4x 10"^ gm
Number of fusions in war head
# Illustrative Example 4.26
10.5x10'^
A reactor is developing nuclear energy at a rate of
32,000 kilowatts. How many kg of
undergo fission per
second ? How many kgof would beused up in 1000 hours ofoperation ?Assume an average energy of200MeV released per fission. Take Avogadro's number as 6 x
and
lMeV=1.6x 10"j.
(200xl0^)(1.6xl0"^^) ^ =>
V=0.1279kg
# Illustrative Example 4.28
Calculate theenergy released bythe fission of 2 gm ofin kWh. Given thatthe energy released per fission is 200MeV.
Solution
Solution
Power developed by reactor E = 32,000 kW
^
The number ofatoms in 2gm of^^jU are
E = 3.2 X 10'watts
A
2x6.025x10^^
235
atoms
Energy released by rector per sec in eVis
p.J:^MeV/s 1.6x10"'^
Energy released per fission V=200 MeV
Pf=2x lO^OMeV/s
=>
A = 200x 1.6x io>3j
As in eachfission even200 MeVenergyisreleased, sonumber offission events occurring in the reactor per second are
=>
7^=3.2 xlO"j
„
2x10^"^ 200
"
Energy released by 2 gm o f i s
^ 2x6.025x10^^ . ,, E= 235 x(3.2xl0ii)j
Nuclear Physics and Radioactivity
146
^ 2x6.025x10 23 235
# Illustrative Example 4.30
(3.2x10"") 3600x10^
kWh
In a nuclear reactor, fission is produced in 1 gm of
(235.0349 amu) in assuming that j^Kr (91.8673 amu) and "^Ba
= 4.55 X lO^kWh. ^Illustrative Example 4.29
In neutroninduced binary fission of^92^ (235.044 amu) two
(140.9139amu) are producedin all reactionsand no energy is lost, write the complete reaction and calculate the total energy produced in kilowatthour. Given 1amu= 931.5MeV. . Solution
stable end products usually formed are42M0 (97.905 amu)and
"JXe usually formed (135.917 amu). Assuming that these isotopes have come from the original fission process, find (i) what elementaryparticles are released (ii) mass defect of the reaction (iii) the equivalent energy released.
The nuclear fission reaction in the above process is
2gU+;n> •> +3> +3'n The sum of the masses before reaction is 235.0439+1.0087 = 236.0526amu
Solution
The sum of the masses after reaction is
(i) The reaction is represented by
140.9139+91.8973+ (1.0087) =235.8375 amu Mass defect,
From the reaction, it is obvious that the total Z value of two
Am=236.0526235.8373 = 0.2153amu
Energy released in the fission of^^^U nucleus isgiven by
stable fission product is (42 + 54) = 96, which is 4 units more
AE = Am X 931.5 MeV
than Z value of left hand side. = 0.2153x931.5^200 MeV
This showsthat the originalunstableproductsmust haveemitted 4 p particles.Again mass number on right hand side is 2 units less than on left hand side i.e., two neutrons are also produced. Now the reaction can be represented as
gu + in ^ fjMo + 'I^Xe + 4„?e + 2 >
Number ofatoms in 1 gm of
are
6.02x10 N=
23 
= 2.56 xlO^'
235
Energy released in fission of 1 gm of £: = 200x2.56x 10^' MeV = 5.12x 10^3 MeV
(ii) Mass defect,
=(5.12x10^)x(1.6x10'3) A/n = mass of L.H.S.  mass ofR.H.S.
Mass of
L.H.S. =(235.044 +1.009) = 236.053amu
Mass of
R.H.S.=(97.905 + 135.9]7)+4(0.00055)
= 8.2xl0'®joule 8:2x10
10
3.6x10^
+ 2(1.009)
kWh
= 2.28x lO^'kWh
=235.842 amu
# Illustrative Example 4.31 Thus mass defect is
Aflj =(236.053235.842)amu = 0.211 amu
The equivalent energy released is
A deuterium reaction that occurs in an experimental fusion reactor is in two stages :
(i) Two deuterium (^D) nuclei fuse together toform a tritium nucleus, with a proton as a byproduct written as D (D, p) T. (ii) A tritium nucleus fijses with another deuterium nucleus to
A£' = Am X 931.5 MeV = 0.211 X 931.5 MeV = 196.54 MeV
form a helium jHenucleus with neutron asa byproduct, written as T(£), n) jHe.
J_Nudear Physics and Radloactiwty
147;
Compute (a) the energy released in each oftwo stages (b) the fission reactor is 10%. To obtain 400 megawatt power from the energy released inthecombined reaction perdeuterium, and(c) power house, how much uranium will berequired per hour? what percentage of the mass energy of the initial deuteriumis (c=3xioWs). released.
Given
2D=2.014102amu =3.016049 amu
'^He=4.002603 amu 
Solution
Powerto be obtainedfiom power house= 400 MW
In this caseenergyobtained per hour= 400MW x 1hour
'H = 1.007825 amu
=(400 X10^ watt) X3600 s
'n = 1.00665 amu
= 144x lO'Oj
Solution
Here only 10% of output is utilized. In order to obtain
(a) The reaction involved in the process is
power house is given by
144 XlO'® joule ofuseful energy, the output energy from the
^D +fD>?T +;p +E, E =
(144xl0*®)xl00 10
Mass defect of the above reaction is
= 144x 10"J
Am = [2(2.014102)(3.016049+1.007825)]
Let, this energy is obtained Ifom a massloss ofAm kg. Then = 0.00433 amu
(Am)c^ = 144x lO^'j
Energy released in the process is £•] =0.00433 x931.5MeV=4.033MeV
or
Aw=
The reaction involved in the process is
144x10"
^
(3x10^)^
= 16x 10 ^kg
^
=0.16gm.
i'T + jD^2^He+Jn +E2 Mass defect of above reaction is
Am = [C3.ai6049 +2.014102)(4.0a2603 +1.008665)]
Since 0.92 milli gram (=0.92 x gm) mass islost in 1gm uranium, hence for a mass loss of0.16 gm theuranium required is given by , , Am
= 0.01888amu
Energy released in the process is
^2 = 0.01888 X931.5MeV=17.586 MeV
1x0.16
= 174gm.
0.92x10"^
Thus torun the powerhouse, 174 gm uranium isrequired per hour.
Total energy released in both processes # Illustrative Example 4.23 = 17.586+4.033=21.619 MeV
(b) Energyreleasedper deuteriumatomis
atmosphere is 2cals/cm^/min on asurface normal to the rays of sun.
21.619
^, = ^—=7.206 MeV (c) % of rest mass of
The energy received from thesun by earth and itssurrounding
released
(a) What is thetotal energy received injoules byearth andits atmosphere.
(b) What is the total energy radiated in J/m by sun to the 7.206
/= 2.014102x931 = 0.385%
universe ? Distance ofsun to earth is 1.49 x 10^ km.
(c) Atwhat ratein megagrams perminute must hydrogen be consumed in the fusion reaction to provide the sun with the
# Illustrative Example 4.32
energy it radiates ?
In the process of nuclear fission of 1 gram uranium; the mass lostis 0.92milligram. Theefficiency ofpower house run bythe
Takemass of hydrogenatom = 1.008145amu. Take mass ofHe atom = 4.003874 amu.
Nuclear Physics and Radioactivity
i148
# Illustrative Example 4.34
Solution
(a) LetD bethe diameter ofearth.Theneffective areaofearth A nuclear reactor generatespowerat 50% efficiency byfission
of2^2^ into two equal fragments of'{^Pd with the emission of
receiving radiation normally is given as
two gamma rays of 5.2 MeV each and three neutrons. The
average binding energies per particle of
and '.J^Pd are
7.2 MeVand8.2MeVrespectively. Calculate the energyreleased in one fission event. Also estimate the amount to
^4^2
71(1.27x10'')' , 2
consumed
per hour to produce 1600megawatt power.
km
=•(1.27)^x iQi^cm^
Solution
Energy released in one fission =Binding energy oftwo 'Ip'd
Energyreceived byearth per minute
nuclei  Binding energy of
=1^(1.27)^xI0'^jx(2x4.2) J/m
nucleus  energy oftwo emitted
gamma rays
Here binding energy of
= 10.645 xlO'«J/m
nucleus is
(A£)y=72 X235 = 1692 MeV
(b) The areaofthe surface surrounding sun at a distance equal Binding energy oftwo '{gPd nuclei to earth distance = 47rc^ 2{SE)p^=2x8.2x116=1902.4MeV
= 4jr(1.49xl0Yx lO^x iC^cm^
Energyof twoemittedgamma raysis
=4p(1.49)^x lO^^cm^
2£^ = 2x5.2=10.4MeV
The energy isreceived atthe rate of2cal/cmVmin on this surfece.
Total energy released is one event is
This amount of energy must be radiated by sun
£=1902.4169210.4
=>
The energy radiated by sun in J/min is
£=200x(1.6xl0'')
£ =2 X4.2 X[471(1.49)2 ^ o26] J/min =>
2.3444 X l(p J/min.
£:=200MeV
£ = 3.2xl0"j
Hence, the number of fission per second required to produce
(c) Weknowthat the fusion reaction in sum is
1600 X10^ J ofenergypersecond (1600 MW) is given as
4H>He + Q N=
1600x10^ 3.2x10
11
= 5x lO'^s"'
The mass defect of this reaction is
So, werequire 5 x lO'^ nuclei of Am=4x 1.008145 4.003874
persecond. Themass of
these atoms will be
=0.028706 amu.
235 m =
»23
x(5xl0'^)
6.02x10'
The energy released in one reaction is
=>
2=0.028706 X931.5 MeV
Thus amount of Now mass of hydrogen required
m= 195.2 X10^gm/3 consumed per hour is
m= (195.2 Xicr^)x3600
4.032580 X1.6598 x 10"^'* x 2.3444 x 10^® ,I9
gm/m
=>
w = 70.27 gm/hr
0.028706x931.5x1.6x10
Since reactor efficiency is 50%, hence consumption of
=3.6673 X10^2 gm/m
= 3.6673 X10'®megagm/m
hour is given by /w=70.27x2 = 140.5gm
per
.Nuclear Physics and Radioactivity
149
# Illustrative Example 4.35
electrical energy obtained fiom the fission of one
nucleus
IS
In the fission of by a thermal neutron, two fission fragments of equal masses and sizes are produced and 4 neutrons are emitted. Find the force between the two fission
fragments atthemoment theyareproduced. Given
1.1 fermi.
£,=3.2xI0"x^=8.0x10'2j. . Therefore, the number offissions of
Solution
7.884x10'^
N=
required peryear is
Since 4 neutrons are produced, the mass number of each
2404 fi"agment (A) = —^— =118. The atomic number ofeach
=9.855 X10^5
8.0x10"'^
Total mass number of^IJPu +neutron (thermal) =239 +1 =240. Number ofmoles of
9.855x10 n
26
= 1.636x 10^
=
6.023 xlO^^
fi"agment = 94/2 = 47.Therefore, charge ofeach fi^agment is q =47 X1.6 X10'^ = 7.52 x lO'^C
required inone year
will be
Therefore, mass of
required peryear is
m=1.636 X10^ X235 =3.844 XI0®g
The radius ofeachnucleus of the fiagment is
=384.4kg
R=R,iAf'
=>
R= lAx l0'5 X(1 18)'/3 (As 1fermi =10"^^cm) # Illustrative Example 4.37
=>
72 = 5.395 X IO'=m
The deuteriumtritium fiision reaction (called the DT reaction)
Distance between the centres of the two fiagments at the
is most likely to be the basic fusion reaction in a future
moment they are produced is
thermonuclear fusion reactor is
fH +5H~>^He +Jn +2
r=2x 5.395x 1015 =>
7 = 10.79x ir'^m
The electrostatic between them is, F =
(a) Calculate the amount of energy released in the reaction,
given ?w (JH) =2.014102 amu, m(]H) =3.016090 amu, m(Jn) = 1.008665 amu andm(2He)=4.002603 amu.
,2
(b) Find the kinetic energy needed to overcome coulomb
4k €o ^2
repulsion. Assume the radius ofboth deuterium and tritium to
1
Q
(7.52xl0y (10.79x10"'^)^
beapproximately 1.5 x 10"'® m. (c) Towhat temperature must the gasesbe heated to initiate the
fusion
reaction?
Take
Boltzmann
constant
A:= 1.38X
=>
F=4.37x10^N Solution
# Illustrative Example 4.36
(a) Theamount of energy released is given by A nuclear reactor using
generates 250 MW of electrical
power. Theefficiency ofthereactor (i.e. efficiency ofconversion of thermal energyinto electrical energy) is 25%. What is the amount of used inthereactor peryear ?Thethermal energy released per fission of
is 200 MeV.
Solution
Rate ofelectrical energy generation is250MW = 250 x 10^W
(or Js~'). therefore electrical energy generated in 1year is (250 X10®Js"') X(365 x24x60 x60 s) =7.884 x 10'® J
2 =[{m (JH) +m(®H)}  {m (^He) +m(^n)}] x931.2 MeV =[{2.014102+3.016090}{4.0092603 + 1.008665}] x931.2MeV = 0.018924 X931.5 = 17.62 MeV
(b) In order to fiise, the two nuclei (D and T) must almost toucheach other. Then the separation between them is equalto twice the radius of each which is given as / = 2/=2x i.5x io'5
= 3.0xl0'®m.
Thermali'energy fiom fission ofone nucleus is200 MeV = In order to startthefusion reaction, deuterium and^tritium have 200x l.6x 10~'^ =3.2x 10""J.Sincetheefficiencyis25%,the to be heated to a very high temperature. At such high
hSO
"
Nuclear Physics and Radioac^vlt^;
temperature thetwo gases areionized, i.e. theonly electron in Wefindtheexcitationenergyfrom the atomic masses. Athermal the atomsofeachgas is removed, sothat the twonuclei havea
neutron hasa negligible kineticenergy (About 0.03 eV).
positive charge of1.6 x10"'^ C, \.e..q^^q2= 1.6 x 10"'^ C. The
potential energy when they are almost touching is
=[m(rt) +
 M(^^^U)]c^
2
=[1.0087amu+235.0439amu236.0456amu]c^ 9X10^ X(1.6 X10"'^)^
=>
U=~
=>
^
3.0x10"'^
=00070 X931.5 =6.5 MeV
^
£'(239u+)= [^(„) + M(23^U)  M(23^U)]c2
(7=7.68 xlO^^J
u*to overcome Coulomb n } u The kinetic energy must bejust»enough
=[1.0087amu+238.0508amu239.0543amu]c^ •'
repulsion between positively charged nuclei. Hence kinetic
=00052x931 5=48MeV
energy must be equal to potential energy calculated above.
Thus we have
Thus KE =7.68 X1o'"^ J
, , , • willjustovercomethe • .L Ifthisenergyissupplied,thetwonuclei repulsion and stay in touch.
has more excitation energy than
when both
are produced by thermal neutron absorption. This iswhy more easily undergoes thermal neutron fission. . ^ # Illustrative Example 4.39
(c) Let The the required temperature of the nucleito initiate
the hision reaction . Then the kinetic energy ofthe system at Calculate, the groundstate Q value ofthe induced fission temperature Tcan be given as reaction in the equation
KE=\kT
«+
Where k is the Boltzmann constant. Hence to start fusion ""
+ dermal. A thermal neutron is in thermal
reaction this kinetic energy will be sufBcient to fuse then equilibrium with its environment; it has an average kinetic together thus for this temperature is given as energy given by 3/2 kT.
Given masses are m («) = 1.0087amu; M( U) = 235.0439 amu;
j= ^37: =3x1.38x10"" 2x7.68x10'^ = 3.7xlO^K
# Illustrative Example 4.38
M(^Zr) =98.916amu;M('^^Te)= 133.9115amu. Solution
Kinetic energy ofa thermal neutron canbeneglected; even for
atemperature of10^ K, the thermal energy isonly 130eV. The Q Calculate the excitation energy of the compound nuclei produced when
^
and
value of the above reaction is given by the equation
absorb thermal neutrons. Given
0=AOTx931.5MeV
masses are
^
11^^235^^ =235 0439 amu, M(n) = 1.0087 amu.
M("^U) =238.0508 amu,
defect ofthe reaction, given as Am = [M ("^U) + m(«)]  [M (^Zr)+ M('^''Te) +3 m(«)]
=[235.0439 98.9165133.91152(1.0087)] amu
M(236u) =236.0456 amu, =0.1985 amu
M("\l) =239.0543 amu Thus energy released is Solution
_
^
2 = 0.1985x931.5= 184.90MeV
The two reactions are
n+
^ ^ 238^
Even with a thermal neutron of negligible kinetic energy a
239y*
tremendous amount ofenergy is released.
^Nuclear Physics and Radioactivity
151
# Illustrative Example 4.40
Substituting given values ofmasses into the expression for 101 gives
Show that
does notdecay be emitting aneutron orproton.
Ie I= (14.014403 140.11180) w.
Given masses are
M
= 230.033927 amu; M
= 229.033496 amu; M
(^^a)=229.032089 amu; m(n) =1.008665 amu, m(p) =1.007825 amu.
=>
ei=0.003223 x 931.5MeV
=>
2l=3.00MeV
Substituting this value Solution
m=M('H), A/=M('X) The corresponding decay equations would be
(i)
intoEquation(4.56), weget:
^55u^« + 22^^and
(ii)
^
1.007825+13.003355
=
+
.
11003355
= 3.00 = 3.23 MeV.
In first equation the energy released can begiven by ^ = [230.033927229.0334961.008665] x 931.5MeV =7.7MeV.
# Illustrative Example 4.42 The nuclear reaction,
Because g < 0, neutron decay is not possible spontaneously.
Similarly insecond equation theenergy released can begiven as
e=[230.033927 229.0320891.007825] x 931.5 MeV =5.6 MeV
As 0 < 0, proton decay is alsonot possible spontaneously.
is observed to occur even when very slowmoving neutrons (M„ = 1.0087 amu) strike a boron atom at rest. For aparticular reaction in which = 0, the helium (M^^ = 4.0026 amu) is observed tohave a speed of9.30 x 10^ m/s. Determine (a) the kinetic energy ofthe lithium = 7.0160 amu), and (b) the ^value ofthe reaction. Solution
# Illustrative Example 4.41
Compute the minimumkineticenergyofprotonincidenton nuclei at restin thelaboratorythatwillproducetheendothermic
reaction '^C(p,
« + '°B^5Li +^2He
Given masses are
(a) Since the neutron and boron are both initially at rest, the total momentum before the reaction is zero and afterward is also zero. Therefore,
M{'^C) = 13.003355 amu, M('H) = 1.007825 amu
We solve this for and substitute it into the equation for kinetic ^ergy. We can use classical kinetic energy with little
m(«) = 1.008665amu,
error, rather than relativistic formulas, because
v^e 9.30 X10^ m/s is not close to the speed oflight cand
M('^N) = 13.005738 amu
will be even less since A/y >
Thus we can write:
Solution A/.:
The thresholdenergyof the incidentprotonsin the lab frameis given by
^Li = m + M
•\Q\
^th = M The magnitude ofthe Q value ofthe reaction is
...(4.56)
Q = '"final 
initial
=[MC^^N) +mJ  [MC^^C) +M('H)]
2A/,
We putin numbers, changing the mass in u tokg andrecalling that 1.60 X10"'^J= 1MeV:
(4.0026)^ (1.66 X10~^^ )(9.3Q x10^) 2(7.0160)(1.66 xlO"^"') A:Li = 1.64x 10'3J= 1.02 MeV
Nuclear Physics and Radioactivity:
•152
(b) We are given the data
(v)
]D +
K^ = K^=0, Q = Kn + K, •He'
so
1
^He 2"^He^e
where
=y(4.0026)(L66x 102^) (9.30X 10^)2
A fiision reaction of the type given below
\T+ Ip +_AE
is most promising for the production of power. HereD and T stand for deuterium and tritium, respectively. Calculate the mass of deuterium in gram required per day for a power output of
lO^W. Assume that efficiency ofthe process to be 50%. Given •2t,
mass of
= 2.87 X10"'^J= 1.80MeV
mass of\T=3.01605 amu.
2 = 1.02MeV +1.80 MeV=2.82 MeV.
Hence,
=2.014102amu.
mass of \p=1.007825 amu. and 1 amu =931.5 MeV.
Web Reference at www.phvsicsgalaxv.com
Age Group  High School Physics  Age 1719 Years
[1778.51 gm]
Section  MODERN PHYSICS
Topic  Nuclear Reactions Module Number  1 to 9
(vi) The binding energies per nucleon for deuteron (^H) and helium (^He) are 1.1 MeV and 7.0 MeV respectively. Calculate the energy released when two deuterons fuse to form a helium
nucleus (^He). Practice Exercise 4.4 [23.6 MeV]
(i) Find the amount of energy produced in joules due to fission of 1gram of uranium assuming thatO.l percent of mass is transformed into energy.
Takel amu=1.66x 10~^^kg = 931.5MeV
(vii)
Find the energy released when
nucleus undergoes
fission by athermal neutron into 3gKr and 'sgBa. Given, mass of^g2U =235.043925 amu, mass ofneutron = 1.008665 amu, mass of3gKr =91.8973 amu and mass of^5^63= 140.9139 amu.
Mass ofuranium = 235 amu.
Take 1amu = 931.5MeV.
Avogadro number = 6.02 x 10^
[200.64 MeV]
[8.978 X 10'° J]
(viii) How much energy is needed to remove a neutron fiom
the nucleus ofj^Ca ?Given
= 1.008665 amu, atomic mass of
(II) What is the power output of reactor if it.takes 30 daysto useup 2 kg offuel, and if eachfissiongives 185MeV
2iCa =40.962278 amu and atomic mass of^JCa =39.962591
of usable energy ?
amu. Take 1amu = 931.5 MeV
[58.52 MW]
[8.363 MeV]
(III) Assuming that 200 MeV ofenergy is released per fission of uranium atom, find the number of fission per second required to release one kilowatt power.
(ix)
[3.125 X 10'^ fissions per second]
amu and 7.016004 amu respectively.
(iv)
It is proposed to use the nuclear fusion reaction
Find the Q value ofthe reaction p + \\^ '^He + '^He.
Determine whether the reaction is exothermic or endothermic.
Theatomicmassof'H, '^He and'Li are 1.007825 amu,4.002603
[Exothermic, 17.347 MeV]
JHh jHj^He in a nuclear reactor of200 MW rating. If the energy fi"om the above reaction is used with a 25 % efficiency in the reactor, how many grams of deuterium fuel will be needed per day. (The
masses of^H and ^He are 2.0141 atomic mass units and 4.0026 atomic mass unit respectively and take 1amu = 931.5MeV).
4.8 Properties of Radioactive Radiations We know that an unstable nuclei achieves stability by radioactivity due to emission of a, P and yradiations. Now in this section we will discuss how these are emitted and the
characteristics of these radiations one by one. [120.31 g]
'Nuclear Physics and Radioactivity
15^
4.8.1 Alpha Decay
and from energy conservation we can say that the released
We've already discussed that the range of attractive nuclear forces amongnucleons isveryshort and that of the electrostatic
energy Q is distributed in both, hence we have ...(4.60)
repulsiveforces betweenprotonsis unlimited. This is the reason
why nuclei with mass number above 210 aresolarge thatthe short range nuclear forces thathold thenucleon together are
Q~ T2 ^ y ^
not sufficient to counterbalance the mutual repulsion between the protons. Alpha decay occurs in such nuclei as a means of increasingtheir stabilitybyreducingtheir size.
2=
IT m
V
1+ m.
m,.
force inthenucleus. The reason isvery high binding energy of aparticles. To escape from nucleus, a particle should have sufficiently high kinetic energy and only aparticle mass is
OT. m.
sufficiently smaller than the masses of its constituent nucleons for such energy to be available.
Theequation ofthenuclear reaction for adecaycanbewritten
a
my +/«£
In this casea question ariseswhyonlyaparticles are emitted why notindividual protons areemitted todecrease therepulsive
Whenever an aparticle is emitted, the mass number of the daughter element is less then 4 and atomic number less then 2.
'a
m.
E

•AA
\Q
...(4.61)
Equation(4.61) shows that the kinetic energy of ejected aparticle in the decay process is
AA
times the gvolunie
of the nuclear reaction in the process.
as
zZjY + 2^6 + energy
...(4.57)
It is seen that all the aemitter radioactive element have mass
number^ > 210 sowe can say that most ofthe energy produced Inequation(4.57) we can see thatduring adecay some energy during decay appears as the kinetic energy of the emitted is released, this shows that the ^value of the adecay is positive and the total released energy is shared by both, the daughter nucleus and the emitted aparticle as shown in figure4.11
aparticle.
4.8.2 Beta Decay
Similar to emission of adecay, pdecay also makes the composition ofnucleus more stable. There are three fundamental beta decay processes are observed. There are
(1) Beta Minus Emission (P"): P" are the electrons,which areemitted from thenucleus when insidea neutron decays to a
(a) Figure 4.11
proton and an electron, this electron is emitted from the nucleus
The energyreleased in the above process or the gvalue ofthe above nuclear reaction can be given as
with some kinetic energy and it is called P" decay, theprocess involved insidethe nucleus in p" decay is + v.(P emission)
...(4.58) Here
m'X
In reaction represented by equation(4.62) also includes a particle with P" emission it is v (nubar), denotes an
mass ofparent nucleus X
m.
> mass ofdaughter nucleus Y
m.
mass ofaparticle (^He)
antineutrino. About this we'll discuss in next section.
During aemission linear momentum of system must be conserved, hence if the speeds attained by aparticle and the daughter nuclide Yin the aboveprocessare and then we must have m
V
y y
=m V a
a
...(4.62)
...(4.59)
(2) Beta Plus Emission (p"'"): P"'^ are the positions, which are
emitted from the nucleus when inside a nucleus a proton transforms to a neutronand position. The position emitted from the nucleus, is called P"*" decay. Theprocess involved inside the nucleus in p'^decay is » « H
I V(P"^ emission)
...(4.63)
Nuclear Physics and Radioactivity ,!
i154
In the above equation(4.63) the reaction involves another (2) Law of Consen^tion of Linear Momentum particle v (nu)with emission, called neutrino. About thisalso During pdecaywhen the directionsof emitted pparticles (e~ we'll discuss in next section. or e^) andthat ofrecoiling nuclei areobserved, theyare almost When a proton inside a p"^ decay inside a nucleus, the parent never exactlyoppositeas required for linear momentum to be radionuclide Xtransforms to a daughter nuclide according to conserved as shown in figure4.12. It seems that one more particle is to be emitted in other direction to conserve linear the reaction momentum but no other particle was detected during beta A Z1
\X
/
+
+
V
...(4.64)
emission.
Here position emission leads to a daughter nucleus of lower atomic number Z by 1 and leaving the mass number A unchanged.Also remember that a proton can not decayinto a
Parent Nucleus
neutron outside a nucleus because of its smaller mass. This is
possible only inside a nucleus.
(3) Electron Capture or Kcapture: This is a third kind of
Pdecay itproperties are similar top"^ emission inwhich anucleus
p particle
pulls in or capturesoneof the orbitalelectronsfromoutsidethe nucleus and inside the nucleus a proton absorbs this electron and transforms into a neutron. This process is called electron
Daughter Nucleus
capture, or Kcapture, sincethe electron is normally captured Figure 4.12
from the innermost or Kshell. The process ofelectron capture inside a nucleus can be given as
(3) Law ofConservation of Angular Momentum \P
r
> « + V
...(4.65)
As in this process also a proton transforms into a neutron, neutrino is also emitted and we can say that basically this
process resembles with p"^ emission but no particle or
are emitted from the nucleus. For a radioactive
Inside a nucleus, every nucleonneutrons protons, electrons or
positionseach have a spin angular momentum± • If we once again look into the equations of Pdecayprocesses, these are
element Xdecays to a daughter element Y, by electron capture, p decay n
the reaction can be given as X
> zxY+v
...(4.66)
A 1 momentum . ±2 Angular
Here also similar to p"^ decay the atomic number of daughter nuclei decreases by one and mass number remain unchanged. 4.8.3 Apparent Violation ofConservation Laws in bdecay As we've discussed that by beta decay also the nucleus becomes more stable. During beta decay it is observed that the basic conservation laws ofenergy, linear momentum and angular momentum are all apparently violated as discussed now.
P"^ decayp
...(4.67)
p + e
^± 2 \ h ^ 2 \ 2^ h ...(4.68)
n + e*
Angular momentum ± 2 2^ 2 2^ ^ 22i In both of above equations we can see that when a nucleon decays in the two on left hand side of equation before decay its
angular momentum is either +
or
but afterdecay
on right hand side ofequation the total angular momentumwill (1) Law ofConservatiMi ofEnergy
be±
iTt
or 0.
We know when a radionuclideXdecays to a daughter nuclide
Twith Pdecaythe amount of energy released or value of the nuclear reaction can be calculated by using mass defect ofthe reaction. But when the emitted Pparticles are observed they do not have enough energy to account for the total energy released. If pparticle carries away only a part ofthe energy, where the remaining energy is going. This was a serious puzzle among physicists in late 1920 s. It was very difficult to accept that law of energy conservation is not valid in this case.
Thus here we can see that on the two sides ofthe equation(4.67) and (4.68), angular momentum are not equal. For an isolated system like nucleus, we can not expect this to happetl. But again this one can not accept that the law of conservation of angular momentum is not applicable here. The above three conservation laws apparently violet in the
process of pdecay. This remain a series problem among
'Nu(^ar Physics and Radioactivity
155
physicists until 1930 when Pauli proposed that during pdecay 4.8.5 Mass Defect Calculation For bdecay one more particle is emitted, called neutrino. Existence of
neutrino wasverified experimentally in 1956. Letsdiscuss Pauli's Neutrino Hypothesisin detail.
We've discussed that there are three fundamental Pdecay processes. Here mass defect involved in .the nuclear reaction
are very small as mass of Pparticle (e~ or e^) is very small. Generally the masses we use for calculation ofmass defect in a
4.8.4 PauIPs Neutrino Hypothesis
nuclear reaction are themasses ofatoms where weneglect the
As discussed above Pauli suggested that during emission of a
masses ofelectrons which is included in the mass ofatom.
pparticlean additional particleisalsoemitted, namedneutrino and this hypothesis made all the above conservation laws
working finely for the process of pdecay. According to Pauli the particleneutrino has the following characteristics:
Here for the Pdecay process the mass of pparticle itself is sameas that ofan electron, we can not neglectthe weightofall
the orbital electrons present inan atom thus for an atom ^^ Z
(i) Its charge is zero
(mass of electron) thus for calculation of mass defect for a Pdecay process, we first find the mass ofnuclein involved in
(iO It is an energy particle like a photon.
the reaction then wefindmassdefect. Lets discuss this one by
(iii) Its rest mass is zero like a photon.
one for each type of Pdecay.
(iv) It carrieslinear momentum like a photon.
(i) Mass Defect for P" Decay:
(v) It carries angular momentum like other nucleons ±
1 h
.
2 27c
(vi) During Pdecay, p^emission is accomplished byemission
For an element
when undergoes p~ emission, the decay
equation is written as
of a neutrino (v) where as P"emission is accomplished by •r'fiY +' I ^te + \
!a
emission ofan antineutrino (v )
(vii) It is extremely difficult to detect a neutrino because it interacts very weakly with matter. It can be concluded on basis of experiments that an average neutrino can penetrate about lO'^ km distance in the most dense material lead without
interacting with it.Even though more then 10^ neutrinos pass through our bodyeverysecond withoutany effect. It isdifficult
' Z+l'
Here the mass defect of the above reaction can be written as
Am =
ZmJ ((Mj. (Z+ 1)
+ mj
or Am = M^ My Thus 2value of this reaction can be given as
AE^{M^My)C^
but in Japan Super Kamiokande neutrino detector was made to detect neutrino.
(ii) MassDefectof P"^ Decay: Thus the new reactions for pdecay processes can be rewritten as
For an element
when undergoes P"^ decay, the decay
equation can be written as
P~decayrt
> p + e~
P^ decayp
>n +
\
..".(4.69)
zJ, + ^^e + V
:a
+w
... (4.70) Here the mass defect of the above reaction can be written as
Now using the above two equations if we again look into the
Am ={Mj^Zm^)
energy, momentum and angular momentum of the nucleons
involved in the decay process all conservation laws now seems to appear valid due to the presence of this additional particle in the reaction.
or
mj
AmMyMy2m A ' y e
Thus 2value of this reaction can be given as
AE =(Mj^MY2m)C^ Herewe'vediscussed that the emittedPparticles carriesenergy lessthen that evolved in the process. Theremainingenergy is (iii) KCapture: carried by the neutrino. Thus we can say that for every emission
the sum ofenergiesof PparticIe and neutrino(or antineutrino) remains constant in the decay.Similarly the direction of emitted neutrino will account for the conservation oflinear momentum
and in the equations for decay processes, if we account for the angular momentum ofneutrino (or antineutrino) then on botli side of equality angular momentum will get conserved.
Foran element
undergoes Kcapture, the decayequation
can be written as
lA
Here the mass defect of the reaction can be given as
Am = (M^ Zm^)  (My (Z1) m)
Nuclear Physics and Radioactivity {
•J56 We know that
or
Thus gvalue ofthe above reaction is given as
AE =
My
(i) The emission of aparticle (jHe) decreases the charge number by two and mass number by four.
C?
Thus emissionof a, aparticles reduce the charge number by
4.8.6 Gamma Decay
2a and mass number by 4a. Just like an atom, nucleus can also exist in different energy states higher then its ground state. An excited nucleus is
denoted by an asterisk after its symbol like Such excited nuclei return to their ground state byemitting photons ofenergy equalto the energydifferencebetween the final and initial states in the transitions involved. These photons have range of energies upto several MeV are traditionally known as gamma rays (yrays).
Whenever a radionuclide imdergoes a or pemission, the daughter nuclide may remain in excited state just after decay which after some time (its life time) transit to ground state with release ofyray photon.
(ii) The emissionof Pparticleincreasethe chargenumber by one and leaving the mass number unchanged. Thus emission of h, Pparticles increase the charge number by 6x1 = 6. Thus
Applying the law of conservation of charge number and mass number before and after the decay, we have 92 = 82 + 2o6 and
Lets consider an example Pactive Mg which decays to Al. Look at the following reactions: Mg
>A1* + P + V
... (4.71)
Al*
>A\ + Y
...(4.72)
In the above reaction we can see when P" is emitted from Mg nucleus some part of the released energy is absorbed by the daughter nucleus & gets in excited state Al* and remaining
energyis carried by p" & v .After a short durationAl* releases the excess energy in form of yray photon and transit to its ground state Al as shown in equation(4.72). Some times due to moreexcitatioii energyoneor may be twogamma decays takes place for daughter nucleus to reach the ground state. Some times the energy ofyray photon from the nucleus which it releases at the time oftransition from higher state to groimd state, is absorbed by the atomic electrons around it. This process is similar to photo electric process in which this atomic electron is emitted with a kinetic energy equal to the nuclear excitation energy minus the binding energy of electron in the atom. This phenomenon we call internal conservation.
U^^gPb + adHe) + 6C?P)
92'
238 = 206+4a
How many alpha and beta particles are emitted when uranium
(^92^) decays to lead Solution
Let a and b, the number ofa and p particles are emitted when
decays to ^"^Pb.
...(4.74)
From equation(4.74), a = \
From equation(4.73), 6 = 1
# Illustrative Example 4.44
Thenucleus ^^Ne decays bypemission into thenucleus ^^Na. Write down the Pdecay equation and determine the maximum
kinetic energy of the electrons emitted. Given m (foNe) = 22.994466 amu. and m(f^a)=22.989770 amu. Ignore the mass ofantineutrino (v). Solution
The pdecay of^^Ne is represented bythe equation
foNe>55Na +e+V+e Mass defect for the above reaction is
Am=[Oe)M(fjNa)] # Illustrative Example 4.43
...(4.73)
= [22.99446622.989770] amu = 0.004696 amu
Now Qvalue of the above reaction is given as e = Amx931.5MeV = 0.004696x931.5 MeV =4.374 MeV
^ _ 157j
INuclear Physics and Radioactivity
The energy Q released is shared by ^^Na nucleus and the (a) Find the values ofA and Z in the above process. electronantineutrino pair. Since ^^Na is massive, most ofthe (b) The alpha particle produced in the above process is found kinetic energy is carried by the e  v pair. The electron will
carry the maximum energy ifthe antineutrino carrieszero energy. Thus the maximum energyof the electrons emitted is 4.374 MeV.
to move in a circular track ofradius 0.11 m in a uniform magnetic field of3 Tesla. Find the energy (in MeV) released during the process and the binding energy of the parent nucleus X. Given that:
# Illustrative Example 4.45 m (Y) =228.03 amu;
A radioactive source in the form ofmetal sphere of diameter
m(Qn) = 1.009 amu
1memits beta particleataconstant rate of6.25 x 10'® particle
m(^He) =4.003 amu;
per second. If the source is electrically insulated how long will it take for its potential to rise by 1.0 volt, assuming that 80% of the emitted beta particles escape from the source ?
m(jH) =1.008 amu. Solution
Solution
Let t be the time for the potential of metal sphere to rise by one volt. Then upto this time pparticles emitted from sphere are
Given nuclear reaction is
9^X^^^Y +a 9^X^228^Y +^He
A^=(6.25xl0'®)xf Number of PparticIesescaped in this time are
(a) According to principle of conservation of charge number, we have for above equation
iV^ =(80/100)x(6.25xlO'0)r = 5xlO>^'f
92 = Z+2 1
o
=>
Thus charge acquired by the sphere in t sec
2 = (5xl0i«f)x(l.6xi0'^ = 8 X10'^ / coulomb
... (4.75)
Z=90
And according to principle to conservationof mass number, we have for above equation ^=228 + 4 = 232
{Qemissionof pparticle leadsto a charge e on metal sphere) =>
A =232
The capacitanceC of a metal sphere is given by (b) For circular track ofa charge particle in uniform magnetic
C = 47i€Qxr
field, we have
fUxlO^J ^ 1" 10
\
/
farad
...(4.76)
Here momentum of aparticle is given as
g=Cx7(Here r=lvolt)
p = mv= 2e,x 5 x /•
/j =2 X(1.6 Xl(r'9)(3) (0.11) kgm/sec
•>12'
=>
(8x109)/ =
= qvB
mv = qBr
12
18
We know that,
.2 mv
2
10'
18
xl
Solving it for t, we get
=>
p = 1.056 X10"'^kgm/sec
K.E. ofaparticle is given by / = 6.95 p sec E
# Illustrative Example 4.46
=
(1.056x10"'^)^ 2x4.003x1.66x10"^"^
= 0.0839 Xicr" joule A nucleusX, initiallyat rest,undergo^ alphadecay according to the equation.
g^X^^^Y +a
^ll
0.0839x10"
1.6x10"*^ = 5.244 MeV
MeV
Nuclear Physics and Radioactivity'
:158
The parent nucleus is at rest and hence its momentum is zero. Now the sum ofmomenta ofdaughter nucleus Tand aparticle must be zero i.e., the momentum ofdaughter nucleus 7 is equal and opposite to momenturn ofaparticle.
Mass defect of this reaction can be given as
Am = [M (2^; Po)  M(2°^ Pb)  M(^He)] Am = 209.98264  205.974404.00260 = 0.00564 amu
Thus K.E. ofdaughter nucleus 7is .2
7 = £v=
^ 2My
Thus
(1.056x10"'^)^
value of the nuclear reaction is
e=0.00564 X931 MeV= 5.25MeV
2x228.03x1.66x10"^'^
= 0.001473 X10"joule=0.092 MeV In above reaction, the total energy released is in the form of kinetic energy ofparticles hence we have
=>
. =5.25 X(1.6 xl0'3)
^
=8.4 XI0~^^joule
The decay constant Xis given byfor ^'^o is given by
Ej =E^+E^= 5.336 meV
Total energy
,
Aw = Mass of
0.693
0.693
= 138.6days =0005day
Mass defect of the reaction is
(Mass of 7+ Mass of aparticle)
Let Mgm be required toproduce 1.2 x ]o'joule perday. Number ofnuclei in A/gm
amu= A/^(228.03+4.003)
N=
(6xlO^^)A/ 210
=(232.0387) amu
Now mass defect of parent nucleus
is given as
Now decay rate by N nuclei per day or the activity ofN nuclei are
=92m^ +(23292)w„M^
dN dt
=(92 X1.008+140 X1.009 232.0387)
0.005 X
•
= 1.9573 amu
per day
... (4.77)
Thus energy produced per day
=> Binding energy of parent nucleus X is
(A£);^.= 1.9573 X931.5MeV
= XN
= 0.005XX(8.4X 10")J
...(4.78)
Thisshould beequal to 1.2 x IO'joule.
= 1823.225 MeV.
Hence 0 . 0 0 5 x (8.4 x ir'^)= 1.2 x IQ^ 210
# Illustrative Example 4.47 =>
A/= 1 gm.
... (4.79)
Polonium (^g^Po) emits ^He particles and is converted into lead Theefficiencyis 10%.Hencethe quantityrequiredwill be 10 gm. (^gjPt)). This reaction is used for producing electric power in a space mission ^'®Po has half life of 138.6 days. Assuming an Fromequation(4.77), activityis efficiency of 10% for the thermoelectric machine, how much
^'®Po is required to produce 1.2 x 10^ J of electric energy per
4 = 0.005 X
(6x10^^)10 210
day at the end of 693 days. Also find the initial activity of the material.
=>
(Given :masses ofnuclei M(^'''Po) =209.98264 amu, M(^®^Pb)
Let.4Q be the activitybefore693 days,then from decayequation
=205.97440 amu, M(^He)=4.00260 amu)
yl = y X10^'per day
we have 693
Solution
=
(2) 1386
The nuclear reaction for the above process can be expressed as
mPo
2gPb + ^He
Aq=
j ^32=4.57 X102" perday.
Nuclear Physics and Radioactivity
""
#Illustrative Example 4.48
"
;
First we will determine the disintegration energygofequations
~
"
(4.80)and(4.81):
Find whether alpha decay or anyofthe beta decay areallowed
for ^ggAc. Given masses are
M(^2^Ac) =226.028356
(» =226.017388amu.
,
M
M(2^Fe) =54.938298 amu,
=222.017415 amu, M
M('Ra) =226.025406amu,M{,^He)=4.002603amu.
M(gMn) =54.938050 amu and
=0.000549a™a
For ^ decay the gvalue ofreaction is
Solution
e=[54.93829854.938050
Our first step will be to write the reaction, then to find the
2 x0.000549] ^931.5 MeV
disintegration energy g.lfg>0,the decay is allowed. Alpha decay: ^oqAc> ^slpr + a
'
=_079MeV
xt . Negative value of^ implies thatpositron decay IS notpossible
Q= [Mfi^Ac)  M(2gFr)  M(^He)]c^
spontaneously For electron capture the gvalue of reaction is
=5.50 MeV(Alpha decay is allowed)
e=[54.938298 54.938050] x 931.5MeV
p" decay: ^Ac =^^JXh + p" + v
=0.23 MeV
2=[Mfi^c)MOh)]c2
D
this case.
= 1.12MeV (P~ decayis allowed)
P^ decay: ^^^c
.
Positive valueof Q implies that electron capture is possible in
^^a + p^ + v
#Illustrative Example 4.50
Q [Nl( g^Ac) M(^gpla) —2mJ
Asample of'®F is used internally as amedical diagnostic tool
=0.38MeV (P decay is not allowed) Electron capture : ^J^Ac + e"^ ^^^Ra + v
to look for the effects of thepositron decay {T,^ = 110 min).
How long does it take for 99% ofthe'¥to decay ? ^
.
Solution
e= [M(™Ac)  M(™Ra)]c2 Radioactive decay equation is
=0.64MeV (Electroncaptureis allowed)
In above analysis it is clear that during adecay, gvalue is maximum thus chances ofadecay are maximum.
.
}ct,Y daughter product is plutonium (Pu). The grand daughter (A) a,p,Y (O Y,a,P P) a,Y,P, product can be expressed as : ' (A)
(B)
(Q 239pu
(D)
438 Graphiteand heavywater are twocommon moderators
241 Pu 94ru
used in a nuclear reactor. The function of the moderator is :
431 The halflife of a radioactive substance depends upon : (A) Its temperature (B) The external pressure on it (Q The mass of the substance
P) The strength of the nuclear forcebetweenthe nucleonsof its atom
(A) Toslowdown the neutrons to thermal energies p) To absorb the neutrons and stop the chain reaction (Q To cool the reactor
P) To control the energyreleased in the reactor 439 Alphaparticlesare fired at a nucleus. Whichof the paths shown in figure4.14 is not possible ?
432 The halflife period of a radioactive elementXis same as the meanlife time ofanother radioactive element Y. Initially both
Nucleus
ofthem have the same number ofatoms. Then:
(A) XandThavethesamedecayrateinitially (B) X and Ydecay at the same rate always (Q Twilldecayatafasterratethan.^ p) ZwilldecayatafasterratethanT
Figure 4.14
(A) 1
433 The radioactive decay ofan elementA"toelements Yand K is represented by the equation Ay
Ay
2+r ^
A+4 y
zi ^
^
440 The decay constant X, of a radioactive sample: (A) Decreases with increase of external pressure and
AA y
zi ^
The sequence of the emitted radiations is: (A)a,p,y P) P,a,Y (Q Y,a,P (D) P.Y'Ct
434 The nuclear reaction 'JJCd >
temperature
P) Increases with increase of external pressure and temperature
can occur with
the:
(A) Electron capture (Q Proton emission
P) 2 P) 4
(Q 3
p) Positron capture p) aparticle emission
435 Two radioactive elements Xand Thave halflife times of
50 minutes and 100 minutes, respectively. Samples X and Y initiallycontain equal numbers ofatoms.After 200 minutes, the ratio
(Q Decreases with increase of temperature and increase of pressure
p) Is independent of temperature and pressure
441 When high energy alpha particles (jHe) pass through nitrogen gas, an isotope of oxygen isformed withthe emission ofparticles namedx. The nuclear reaction is
i^e ^ 'JO + x What is the name ofx ?
(A) Electron (Q Neutron
number of undecayed atoms of X . IS number of undecayed atoms of Y
(A) 4
P) 2
(Q 1/2
P) 1/4
442 When boron ('jB) isbombarded byneutron, aparticle is
436 When a P"particle is emitted from a nucleus,^the neutronproton ratio: (A) Is decreased (Q Remains the same
P) Proton p) Positron
p) Is increased p) First (A) then (B)
emitted. The resulting nucleus has a mass number: (A) 11 P) 7 (Q 6 p) 15 443 Consider aparticles, pparticles and yraj^, each having an energy of 0.5 MeV. In increasing order ofpenetrating powers, the radiations are :
437 In the given reaction : AA Z1
K
A~A y Z1 ^
(A) a,p,y (Q p,Y,a
•

P) a,Y,p p) Y,p,a
,
;Nuc[ear"iPhysics,and, Radioactivity
167
444 Aradioactive nucleus emits a beta particle. The parent 448 Figure 4.15 shows aroughly approximated curve ofbinding and daughter nuclei are:
(A) Isotopes (C) Isomers
energy per nucleon with mass number. W, X, Yand Z are four
(B) Isobars (D) Isotones
nuclei indicated on the curve. According to this curve, the process that would release energy is ;
445 Fast neutrons can easily beslowed down by: (A) The useof lead shielding (B) Passingthem through water (Q Elasticcollisions withheavynuclei P) Applying a strong electric field
446 A nucleus ruptures intotwonuclear parts, which have their velocityratio equal to 2 :1. What will be the ratio of their nuclear size (nuclear radius)?
(A) 3"^A
(B) 1:3"2 Figure 4.15
447 Fornucleiwith^> 100,marktheINCORRECTstatanent;
(A) The bindingenergypernucleon decreases on the average as A increases
(B) If the nucleus breaks into tworoughly equalparts,energy is released
(Q Iftwonuclei fiise to form a bigger nucleus energyisreleased P) The nucleus with 2> 83 are generallyunstable
(A) Y^2Z (Q W>2Y
P) W^X + Z P) X^Y + Z
Nuclear Physics and Radioactivity i
i168
NumericalMCQs Single Options Correct 41 Two identicalsamples (samematerialand sameamount)P
andQofa radioactive substance having mean life Tareobserved
(B) 10:11 P) 81:19
(A) 19:81 (C) 15:16
to have activities Ap & Ag respectively at the time of observation. If P is older than Q, then the difference in their
48 Halflives of two radioactive substances A and B are
ages is:
20 minutesand 40 minutesrespectively. Initiallysamples^ and B haveequal number ofnuclei. After 80minutes, theratioofthe remaining number of^ andP nuclei is: (A) 1:16 P) 4:1
(A) Tin
A.
f
(B) Tin .
A. \
N
(C) j In
(0 1:4
(D) T
P) 1:1
A 49 A100 ml solution havingactivity 50 dpsiskeptin a beaker.
42 Bihasa halflife of5 days. Whattime istaken by(7/8)*^ part It is now constantly diluted by adding water at a constant rate of the sample to decay ? (A) 3.4 days (Q 15 days
of 10 ml/sec and 2 ml/sec ofsolution is constantly being taken out. Find the activity of 10 ml solution which is taken out, assuming halflife to be effectivelyvery large :
(B) 10 days (D) 20 days
1/2
1/4
43 A certain radioactive substance has a half life of 5 years.
(A) 10
P) 10
Thus for a nucleus in a sample of the element, probability of decay in 10 years is : (A) 50% (C) 60%
44
A
1/2
1/4
(B) 75% P) 100% >,1
B
X2
^
(Q 50
P) 50
''f
410 A radioactive source has a half life of3 hours. A fi^eshly
C
prepared sample of the same emits radiation 16 times the
^=0
No
0
0
t
N, •]
N. "2
N, "'3
In the above radioactive decay C is stable nucleus. Then : (A) rate of decay of A will first increase and then decrease P) number of nuclei of B will first increase and then decrease
(Q if ^2 > X.,, then activity ofB will always be higher than activity ofA
p) ifXj» Xj, then number ofnucleus ofCwill always be less' than number of nucleus ofB.
45 The halflife of a radioactive substance is 10 days. This
permissible safevalue. Theminimum timeafterwhich itwould be possibleto work safelywith the source is : (A) 6 hours (Q 18 hours
p) 12 hours p) 24 hours
411 The halflife period ofa sample is 100 second. If we take 40 gram of the radioactive sample, then after 400 second how much substance will be left undecayed ? (A) 10gram p) 5 gram (Q 2.5gram p) 1.25gram
means that:
(A) The substance completely disintegrates in 20 days P) The substance completely disintegrates in 40 days (C) 1/8 part of the mass of the substance vvill be left intact at
412 The kinetic energy ofa 300 K thermal neutron is: (A) 300eV P) 300eV (Q 0.026eV p) 0.026MeV
the end of40 days
p) 7/8 part of the mass of the substance disintegrates in
413 A samplehas two isotopes ^^^A and ^B having masses
30 days
50 g and 30 g respectively. A is radioactive and B is stable. A decays to by emitting a particles. The halflife of.B
> C (stable)
At a certain time, the activity of nucleiA is 'x' and the nett rate
the energy Q released is : (A) IMeV (Q 23.8MeV
(B) 1.44sec
(2)(Q 0.69
2H + 2H
(B) 34.7min (D) 1hour
ofincrease ofnumber ofnuclei B is
Theactivity ofnuclei B
(B) 11.9MeV
at this instant is :
P) 931MeV
(A) y
(B) x;
(Q yx
p) X
419 A radioactive substance is beingproduced at a constant rate of 200 nuclei/s. The decay constant of the substance is ls~'. After what time the number of radioactive nuclei will become 100. Initiallythere are no nuclei present: 1
(A) 1 sec
(B)
(Q In (2) sec
(D) 2 sec
In (2)
sec
426 The activity ofa radioactive sample decreases to one third of the original valuein 9 years. Whatwill be its activity after further 9 years ?
(A) ^(/3
(B) A^6
(Q A^9
(P) A^IS
427 The radioactivity ofasampleis^atatime/,and Tata time/jIfthemean lifeofthe specimen isx,thenumber ofatoms
420 What is the rest mass energy ofan electron ? (B) 0.51 MeV (A) leV (Q 931MeV (D) None of these
that have disintegrated inthe time interval (/^  /j) is: (A)X/,y/2 (Q iXY)/x
(B) XY p) (XY)x .
Nuclear Physics and Radioactivity ;
428 A radioactive nucleus can decay by either emitting an a
particle orby emitting a p particle. Probability ofa decay is75% whilethatofpdecayis25%.Thedecayconstantofa decay is X. .
X, and thatof p decay is a.^. — is: (A) 3
(B) Y
(Q 1
P) cannot be said
435 Therateofdisintegration ofaradioactive substance falls from 800 decay/min to100 decay/min in6hours. The halflife of the radioactive substance is ;
(A) 6/7hour (Q 3hrs
(B) 2hrs P) Ihr
436 The activity of a radioactive sample decreases to one
tenth of the original activity in a periodof oneyearfurther. After further 9 years, its activitywill be:
(B) fA
A
429 Part ofthe uranium decay series is shown U238 . 92
90^"
(Q
10
P) None of these
10
„Ra226
Th^^o
90
Howmany pairs of isotopes are there in the above series' (A) 1 •" (B) 2 ^ (Q 3
• P) 0
437 Halflife of a radioactive elements is 12.5 hour and its
quantity is256 gm. After how much time itsquantity will remain Igm?
(A) 50hrs (Q 150hrs
p)' 100hrs P) 200hrs
430 A radioactive element^with halflife of2 hours decays
givinga stable element 7? Afterhowmanyhoursthe ratioofX 438 An unknown stable nuclide after absorbing a neutron to Twillbe 1:7?
(A) 6 hours (Q 10hours
p) 8 hours , . P) 12hours
431 The energy released by the fission ofone uranium atom is 200 MeV. The number of fissions per second required to produce 3.2 MW ofpower is :
(A) 10^' (Q lO's
P) 10^'' P) 10'^
emits an electron, and the new nuclide splits spontaneously into two alpha particles. The unknown nuclide is: (B) 3Li' (A) Pe' (D) cB 10 (Q 439 A radioactive material is made at the constant rate of
10'^ nuclei per second andthe material is gettingdecayed with decay constant 0.0123 month~k Activity ofmaterial after a very longtime (ifinitiallytherewasno radioactive material)is: 10^
dps
P) l.bxio^dps
432 Nuclei X decay into nuclei 7 by emitting a particles. Energiesof a particle are foundto be only 1 MeV& 1.4MeV. Disregarding the recoil of nuclei 7. The energy of y photon
(Q lO^'dps
emitted will be:
440 The halflife ofa certain radio isotope is lOminutes.The
(A) 0.8MeV (Q IMeV
•
0.0123
p) None of these
P) 1.4MeV
number of radioactive nuclei at a given instant oftimeis 1Ol
P) 0.4 MeV
Then the number ofradioactive nuclei left 5 minutes later would be:
433 An elementXdecays, first by positron emission and then
twoaparticles are emitted in successive radioactive decay. If
(A) ^
10
P) lO'^
the product nuclei has a mass number 229 and atomic number 89, the mass number and atomic number ofelement Xare:
(A) 237,93 (Q 221,84
10®
P) 237,94 . p) 237,92
434 The decay constant of a radioactive substance is 1 per month. The percentage of radioactive substance left undecayed after two months will be:' (A) 25% P) 50%
(Q 66%
(Q V2 X10'
' P) 87%
441 A uranium nucleus (atomic number 92, mass number 238) emits an alpha particle and the resultant nucleus emits a
P particle. The atomic and mass numbers respectively of the final nucleus are:
(A) 90,240 (Q 91,234
P) 90,236 P) 92,232
andiRadioactfvity
442 A radioactive nuclide having decay constant Xis 449 In which ofthe following process the number ofprotons produced at the constant rate of n per second (say, by in the nucleus increases. bombarding atarget with neutrons). The expected number A^of (A) a  decay
(B) p" decay (D) kcapture
nuclei in existence tseconds after the number isN^ is given by: (Q P"^decay
(A) N=
(B) N= ^
(Q
+
450 Aradioactive nucleus 'X' decays toa stable nucleus 'Y'. Then the graph ofrate offormation ofA' against time V vdll
+(^0
be:
.
443 The equation
4
>
+ 2e^+ 26MeV represents
(A) pdecay
(B) ^decay
(Q Fusion
(D) Fission
444 At time t=0, Aj nuclei ofdeeay constant A,j &Aj nuclei ofdecay constant ^2 are mixed. The decayrate ofthe mixture at
(A)
(B)
(Q
(D)
timeis:
(A) AiA2e
—(A,] —A2)/
(B)
will be decreasedby a fraction :
(Q. iN{k^e ^i'+A2A.2e ^2')(D) Ajl, 445 The number ofa and p"emitted during theradioactive decay chain starting from (A) 3a&6p(Q 5a&4p
451 Suppose thespeed oflight were halfofthepresent value, theamount ofenergy released in the atomic bomb explosion
Ra andending at (B) 4a&5pP) 6a&6p
Pb is:
(A) i
(s) t
(Q f
(D) I •
452 The graph shows how the countrateA of a radioactive
446 Two identical nuclei A and B of the same radioactive
source varies with time /. A and t are related as : (assume In
element undergo p decay. Aemits a Pparticle and changes to
(12)= 2.5)
A'. B emits a pparticIe and then a yray photon immediately
In A
afterwards, and changes to R': (A) A' and B' havethe sameatomicnumberand mass number p) A' and B' have the sameatomicnumberbut differentmass
\
numbers
(C) A'and 5' have differentatomic numbers but the same mass
\
number
P) A'andR'areisotopes.
10
•
'^N 11
20
30
Time (s)
Figure 4.16
447 In nuclear fission 0.1% mass is converted into energy. How much electrical energy can be generated bythefission of (A) A = 2.5 e^^' Ikgoffuel?
.
(A) IkWh (Q 2.5 kWh
P) lO^kWh P) ZSxlO^kWh
448 Ifthe binding energyper nucleon in jLi and ^He nuclei is respectively 5.60 MeV and 7.06 MeV, then the energy of the reaction
jLi+ Jp(A) 19.60MeV (Q 8.46 MeV
(C) A = 2.5e^^'
453 The ratio of molecular massof tworadioactive substance
is y and the ratio oftheir decay constant is —.Then the ratio oftheir initial activityper mole will be: (A) 2
(B) f
(q
(D) I
2 ^He is P) 12.26MeV P) 17.28MeV
P) A = ne^^' P) A = l2e^^'
Nuclear Physics and Radioactivityj
462 A radioactive element X converts into another stable ofthe a radioactivematerial of halflifeof 30 minutes decreases element Y. Halflifeof^is 2 hrs.Initially only^is present. After time /,theratio ofatoms ofA"and fis found tobe 1:4, then / in to 5 dpsafter2 hours. Theinitial countrate was :
454 Thecount rateofaGeiger Muller counter for theradiation
(A) 80 second"^ (Q 20 second"'
(B) 625 second' (D) 25 second"'
hours is :

(A) 2 (Q between 4 and 6
(B) 4 P) 6
455 A radioactivenuclide can decay simultaneously by two
different processes which have individual decay constants A,,
463 Two radioactive materials A, andAjhave decay constants
and Xj respectively. The effective decay constant ofthe nuclide 10 A, and Arespectively. Ifinitially they have the same number of nuclei, then theratio ofthenumber ofnuclei ofA,tothatofX^
is Xgiven by:
(A)^=V^ (C) X=^i\ +X,)
(B)
+^
will be l/e after a time:
(B)
IOA
(D) A,=A,j + X2
11
(Q
(D)
lOA
1
llA
J_ 9A
456 Anaparticle ofenergy 5 MeV is scattered through 180" bya fixed uranium nucleus. The distance ofclosest approach is 464 A radioactive materialdecays bysimultaneous emission oftwoparticles with respective halflives 1620 and BID years. ofthe order of: (A) lA (B) 10^'Ocm Thetime inyears after which onefourth ofthematerial remains (Q lO'2 cm
P) lO'^cm
is:
(A) 1080 (Q 3240
P) 2430 457 Halflivesoftwo isotopesAand Tofa materialareknown P) 4860 tobe2 X10^ years and4 x 1years respectively. Ifa planetwas formed with equal number of these isotopes, then the current 465 A starinitially has 10^ deuterons.ltproduces energy via
age ofplanet, given that currently the material has 20% ofAand the processes 80% of Tby number, will be:
(A) 2 X10^ years
p) 4 x 1years
(C) 6 X10^ years
p) 8x10^ years
]R+ ^n^]U+p and
>^e + n
458 A nucleus raptures into two nuclear parts which have where the masses ofthe nuclei are : m (^H) = 2.014102 amu, their velocities in the ratio of2:1. What will be the ratio oftheir nuclear sizes (radii) ?
(A) 2'/^: I
P) 1
(Q3"2:1
P)1:3"2
459 In which ofthe followingcase the total number of decays will be maximum in a time interval of/=0 to /=4 hr. The first term
m{p) = 1.007825 amu, m(«) = 1.008665 amu and mCHe) = 4.002603 amu. Iftheaverage power radiated bythe star is.lO'^ W, the deuteron supplyofthe star is exhausted in a time ofthe order of:
(A) lO^s
P) 10®s
(Q lO'^s
P).10 16
represents the number of nuclei at time / = 0 and the second
466 Masses oftwo isobars ^Cuand^n are 63.929766 amu
represents the half lifeofthe radionuclide:
and93.929145 amurespectively. Itcanbeconcluded from these
(A) N^4hr (Q 3No,2hr
(A) Boththe isobars are stable
P) 2No,3hr P)4No,lhr
460 There are two radioactive substances A and B. Decay constant of B is two times that of A. Initially both have equal
number of nuclei. After k halflives ofv4 rate of disintegration of both are equal. The value of«is : P) 2 (A) .l (Q 4 P) All of these
data that:
P) ^Zn isradioactive, decaying to^Cu through pdecay (C) ^Cu is radioactive, decaying to ^Zn through ydecay p) ^Cu is radioactive, decaying to ^Zn through Pdecay 467 A radioactivesubstanceA decaysinto another radioactive
substance Y. Initially only A was present A^ and A^ are the disintegration constants ofA and Y.
and
arethe number of
nuclei ofAand Yat any time /. Number ofnuclei N^. will be 4*61 An atom ofmass number 15 and atomic number 7 captures
maximum when:
analpha particle andthen emits aproton. Themass number and (A)
(A) 14aiid2 respectively (Q17 and6 respectively
(Q
P) 16and4 respectively p) 18and8 respectively
A,
N.
atomic number ofthe resulting atom will be:
A.A,
P)
A A„
P) A^A,=A,A,
A A.
^Nuclear Physics and Radioactiwty
173
468 Aparticle ofmass Matrest decays into two particles of
masses/Kj and/«2, having nonzero velocities. The ratio ofthe (A) — In deBroglie wavelengths ofthe particles, X/ is: ^ (A)
(B)
(Q 1.0
P)
lA.
P) T
•4
T
469 A radioactive nucleus is being produced at a constant rate a per second. Its decay constant is X. If
are the number
ofnuclei attime /=0, then maximum number ofnuclei possible are:
2A,
475 The half life of a radioactive material is 12.7hr. What fraction of the original activematerial would becomeinactivein 63.5hr.
(B) A^o+X
(A) 1/32 (C) 31/32 ,
(D)
476 In the nuclear fusion reaction,
a
(C) A'.
P) nn
li^
p) 1/23 p) 23/32
470 AradioactiveisotopeATwith halflifeof 1.37 XlO^years
given thattherepulsive potential energy between thetwo nuclei
decays to Ywhich is stable. A sampleof rocks from the moon
is~ 7.7 XlO"'"* J, the temperature at which the gases must be
was found to contain both the elements ATand Ywhich were in the ratio 1 :7. The age ofthe rocks is :
heated to initiate the reaction is nearly [Boltzmann's constant A= 1.38 X1023 J/K]
(A) 1.96 X1QS years (Q 4.11 XlO^years 471
(B) 3.85 x 10^ years p) 9.59 x IQ^years
atoms ofa radioactive element emit
(A) lO'K (C) lO^K Particles
per second. The decay constant of the element is (in s"') A^,
(Q A^,ln(2)
(D) iV,ln(2)
p) ID^K P) lO^K
477 The activity ofaradioactive sample is/1, attimef, and^j at time Ifthe mean lifeofthe sample isi, then the number of nuclei decayed in time is proportionalto : {h)A.LAJ,
p)
(Q {A,Ap{t^t,)
P) {A,Apr
472 Sun radiates energy inalldirection. The average energy
578 Nuclei ofradioactive element^ areproduced atrate^2'at
received at earth is 1.4KW/m. The averagedistance between
any time t. The elementA has decayconstant X. Let A^be the
the earth and the sun is1.5 x 10"m. Ifthis energy isreleased by conservation of mass into energy, then the mass lost per day
numberofnucleiofelement^atanytime/.Attime/=/o, ^
bythesun isapproximately (use 1day= 86400 sec)
(A) 4.4x lO^kg (Q 3.8x lO'^kg
p) 7.6x IQi^kg p) 3.8x io"^kg
minimum. Then the number ofnuclei ofelement/^ at time/= is:
•
2/f, —Xtn
473 Power output of reactor ifittakes 30 days touse up 2kgoffuel, and ifeach fission gives 185 MeV ofusable energy, is:
(A)5.85MW (C) .585MW
P) 58.5KW p) None of these
474 A radioactive material of halflife Twas produced in a nuclear reactor at different instants, the quantity produced second time was twice of that produced first time. If now their
present activities are A^ and A^ respectively, then their age difference equals:
(A)
(Q
P)
Iq
2/ft —Xtn
Xiq
P)
479 Suppose aradioactive substance disintegrates completely in 10days. Each dayit disintegratesat a constantrate which is twicethe rate of the previousday. The percentageofthe material
leftto be disintegrated afterpassing of 9 days is : (A) 10 (Q 25
p) 20 ' p) 50
Nuclear Physics and Radioactivity ,
174
AdvanceMCQswith One or More OptionsCorrect 41 When a nucleus with atomic number Z and mass number^
47 Two identical nuclei A and B ofthe same radioactive element
undergoes a radioactive decay process :
undergo similar p decay. Aemitsa pparticle andchanges to.4'. B emits a pparticle and then a yray photon immediately
(A) BothZ andA will decrease, if the process is a decay (B) Z will decrease butA will not change, if the process is
afterwards, and changes to B':
decay
(A) A' and B' may have the same atomic number and mass
(Q Z will increase butA will not change, if the process is p"
number
decay
p) A' andB' mayhavethe sameatomic number but different
p) Zand ^ willremainunchanged, ifthe process is ydecay
mass numbers
(Q A' andB' mayhavedifferent atomic numbers butthe same 42 When the nucleus ofan electrically neutral atom undergoes
mass number
a radioactive decayprocess, it willremainneutralafterthe decay P) 'and 5'may are isotopes ifthe process is : 48 A and B are isotopes, B and C are isobars. All three are (A) An a decay (B) A p" decay radioactive: (Q Ay decay P) A Kcapture process (A) A, B and C must belong to the same element P) A, B and C may belong to the same element 43 Which ofthe following assertions are correct ?
(A) A neutron can decay to a proton only insidea nucleus P) A proton can change to a neutron only inside'a nucleus
(Q It is possible that AwillchangetoB through a radioactive
(Q An isolated neutron can change into a proton p) An isolated proton can change into a neutron
P) It is possible that B willchangeto C through a radioactive
44 Disintegration constant ofa radioactive material is X: logg2 (A) Its halflife equal to —— 1
decay process decay process
49 A nuclide A undergoes adecay and another nuclide B undergoes pdecay: (A) All the aparticles emitted hyA willhave almostthe same speed
p) Its mean life equals to —
P) The aparticles emitted by A may have widely different
(Q At time equalto mean life, 63% of the initial radioactive
speeds
K
material is leftundecayed
p) After 3halflives, yrd ofthe initial radioactive material is
(Q All the pparticlesemittedby5 will have almostthe same speed
p) The pparticles emittedby5 will have differentspeeds
left undecayed.
410 Which of the following is correct for a nuclear reaction?
45 A nucleus undergoes a series of decay according to the scheme
Atomic number and mass numbers ofE are 69 and 172
(A) P) (Q P)
Atomic number ofif is 72 Mass number of5 is 176 Atomic number ofD is 69 Atomic number ofC is 69
(A) A typical fission represented by 92U"^^ + 3gKr^^ +Energy p) Heavywateris usedas moderator in preference to ordinary water
(Q Cadmium rods increase the reactor power when they go in, decrease when they go outward
P) Slower neutrons are more effectivein causing fissionthan faster neutrons in case of
46 The half lifeof a radioactive substance is Tq. At f= 0, the number of active nuclei areN^. Select thecorrect alternative.
(A) The number ofnuclei decayed intime internal 0 Ms N^e'^
411 The decay constant of a radioactive substance is
0.173 (years)"'. Therefore
p) The number of nuclei decayed in time interval 0  / is
(A) Nearly 63% of the radioactive substance will decay in
W„(le^0
(1/0.173) year
(Q The probability that a radioactive nuclei does not decay in interval 0  ds
P) The probability that a radioactive nuclei does not decay in interval I  e~^
p) Halflife ofthe radio active substance is (1/0.173) years. (Q Onefourth of the radioactive substance will be left after nearly 8 years p) All the above statements are true
iNuclear Physics and Radioactivity
175j
412 Which ofthe following statement(s) is (are) correct ?
417 Assume that the nuclear binding energy per nucleoh
(A) The rest mass of a stable nucleus is less than the.sum of (B/A) versus mass number (A) isas shown in the figure4.18. the rest masses of its separated nucleons. Use this plot to choose the correct choice(s) given below. (B) The rest mass ofa stable nucleus isgreater than thesum of the rest masses of its separated nucleons. BIA
(Q In nuclear fission, energy isreleased by fusing two nuclei 8
ofmedium mass (approximately 100 u)
P) Innuclear fission, energy isreleased by fiagmentation ofa 6
very heavy nucleus.
4i
413 Aradioactive samplehas initial concentrationWq ofnuclei (A) The number ofundecayed nuclei present in the sample
2
decays exponentially with time
(B) The activity (R) ofthe sample at any instant is directly proportional to the number ofundecayed nuclei present inthe
100
sample at that time
Figure 4.18
(Q The number ofdecayed nuclei grows linearly with time . P) The number ofdecayed nuclei grows exponentially with time
,
.
200
A
(A) Fusion oftwo nuclei with mass numbers lying in the range of 1
+ a. The atomic masses needed are as ^^P =31.974 amu and that of
=31.972 amu.
follows:
238pu
Ans. [1.86 MeV]
234y
238.CM955amu
234.04095 amu
4.002603 amu
452 The selling rate ofaradioactive isotope is decided by its
Neglect any recoil ofthe residual nucleus.
activity. What will bethesecondhand rate ofa one month old
Ans. [5.58 MeV]
800 rupees?
446 Calculate the gvolume in the following decays:
Ans. [1 87 rupees]
(a)'^O > '^F + e+V
453 In an agricultural experiment, asolution containing 1mole
^(^2"
days) source if it was originally purchased for
(b)^^Al,^ ^^Mg + e^ +v
ofa radioactive material (t^^  14.3 days) was injected into the
The atomic masses needed are as follows :
roots of a plant. The plant. The plant was allowed 70hours to settle down and then activity was measured in its fruit. If the
'^0
i^F
activity measured was 1 pCi, what percent of activity is
19.003576 amu
18.998403 amu
transmitted from the root to the fruit in steady state ?
^^Al 24.990432 amu
25jyjg 24.985839 amu
Ans. [1.26 X lo'i %]
^
Ans. [(a) 4.816 MeV, (b) 3.254 MeV]
454 Calculate the energy released by 1g ofnatural uranium assuming 200 MeV is released in each fission event and that
the fissionable isotope 447 Findthe maximum energythatabeta particle can have in
has an abundance of 0.7% by
weight in natural uranium.
the following decay
,
176lu
F.
Ans. [5.7 X lO^ j]
Atomic mass of "^Lu is 175.942694 amu and that of'^®Hf is 455 A radioactive nucleus X decays to a nucleus Twith a decay constant = 0.1 sec~^ Yfurther decays to a stable 175.941420 amu.
nucleus 2 with a decay constant Xy= 1/30 sec"'. Initially, there are only nuclei and their number is Vg = 10^®. Setup the rate equations for the populations ofX, Fand Z. The population of
Ans. [1.182 MeV]
448 A radioactive sample has 6.0 x 10'^ active nuclei at a
the Ynucleus as a function of time is given by Ny (/) = {®2cp ( Xyt)  exp ( X^f)}. Find the time at
certain instant. How many of these nuclei will still be in" the
which Ny ismaximum and determine the population ofA'and Z
same active state after two halflives ?
at that instant.
Ans. [1.5 X io'8]
[(')
449 The number of atoms in an ancient rock equals the number of ^°^Pb atoms. The halflife of decay of is
ciN^
(iii)A'^= 1.92 X N. = 232 X 10'^]
dN
~jr" ^ =5.76 x IQ'
dN, dl
= X N, (ii) 16.48 s,
4.5 X10^ y. Estimate the age ofthe rock assuming that all the 456 Calculate the 0value of the fusion reaction ^'^^Pb atoms are formed from the decay of ' ''He+ ''He = ^Be.
Ans. [4.5 X lO^y old.]
Is such a fusion energeticallyfavourable ?Atomic massof ^Be is 8.0053 amu and that of'He is 4.0026 amu.
450 Find the energy liberated in the reaction
Ans. [ 93.1 keV, no]
223Ra ^ 209pt, + 14(3
223j^a 223.018amu
20
+ 3"+ v.
Where represents a mercurynucleus in an excited state at energy 1.088 MeV above the groundstate. Whatcan be the maximum kinetic energy of the electron emitted ? The atomic mass of '^^Au is 197.968233 amu and that of
is
197.966760 amu.
to find the wavelength of the Ka Xray emitted following the electron capture. Ans. [(a) Neutrino, (b) 20 pm]
494 A radioactive isotope is being produced at a constant rate dNIdt= /? in an experiment. The isotopehas a halflife Show that after a time t »
the number of active nuclei will
Ans. [0.2806 MeV]
become constant. Find the value of this constant.
489 A nucleusat restundergoesa decayemittingan aparticle
A
ofdeBrogliewavelength,a, =5.76x 10"'^m. Ifthemassofthe daughter nucleus is 223.610 amu and that of the aparticle is 4.002 amu, determine the total kinetic energy in the final state. Hence, obtain the mass ofthe parent nucleus in amu
(lamu = 93I.470MeV/c2) Ans. [6.25 MeV. 227.62 amu]
,
r ^6/2 ,
t 0.693 ^
495 Consider the situation ofthe previous problem. Suppose
the "production of the radioactive isotopestarts at 7= 0. Find the number of active nuclei at time t.
Ans. [yd e^')]
.^Nuclear^Physics and Radioactiwty
496 A body of mass is placed on a smooth horizontal surface. Themass ofthe body is decreasing exponentially with disintegration constant A,. Assuming that the mass is ejected backward with a relative velocity ii. Initiallythe body was at rest. Find the velocity ofit after time t.
; \ 4102 A number
. 185;
of atoms of a radio active element are
placedinsidea closed volume. The radioactive decay constant
for the nucleus ofthis element isA,,. The daughter nucleus that form as a result of the decay process are assuthed to be radioactive toowitha radioactive decay constant Determine the time variation ofthe number ofsuch nucleus. Consider two
Ans. [«>,/]
limiting cases A.j»
497 Radioactive isotopes are produced in a nuclear physics
and A.j« X^.
Ans. [
]
experiment at a constant rate dN/dt = R. An inductor of
inductance 100 mH, a resistor ofresistance 100Hand a battery are connected to form a series circuit. The circuit is switched on
4103 A sample of 100millicurie of krypton gas consists of a
at the instantthe production of radioactive isotope. It is found
mixture ofthe active isotope ^^Kr &the stable isotope ^K.r. If
that i/N remains constant in time where i is the current in the circuit at time t and N is the number of active nuclei at time t.
the volume ofthe mixture is 10 cm^at STP & halflife of^Kr is
Find the halflife of the isotope.
10 years. Calculate the %byweight of®^Kr present inthe mixture. Ans. [0.632 %]
Ans. [6.93 x 10^ s]
4104 Astationary ^g^b nucleus emits an a  particle with
498 A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decayswith an average life t. Find the value of R for which the ratio of the
kinetic energy = 5.77 MeV. Find the recoil velocity of a daughternucleus. What fraction ofthe total energyliberatedin this decay is accoimted for by the recoil energy of daughter
electrostaticfield energystored in the capacitorto the activity
nucleus ?
ofthe radioactive sample remains constant in time.
Ans. [3.4 X iqS mis, 0.020]
Ans. [2t/C]
4105 Energy evolved from the fusion reaction is to
499 A smallbottlecontains powdered beryllium Be& gaseous, be 21H = 2He+ Q usedfor the production ofpower. Assuming radon which is used as a source of aparticles. Neutrons are produced when aparticles of the radon react with beryllium. The yield of this reaction is (1/4000) i.e. only one aparticle out
of4000induces thereaction. Findtheamount ofradon (^^^Rn) originally introduced into the source, if it produces 1.2 x
neutrons persecond after five days. [T,/2 of Rn = 3.8days]
the efficiency of the process to be 30 %. Find the mass of
deuterium that will be consumed in a second for an output of 50 MW.
[Massof ^He =4.002603amu; fH=2.014I02amu] Ans. [2.9 X 10"^ kg]
Ans. [2.1 X 10"^ g]
4106 Findthe amount ofheatgenerated by1.00mgof a ^"'Po
4100 Theenergyofalphaparticlesemittedby^'®Pois5.3MeV.
preparation during the mean lifetime.period of these nuclei, if the emitteda particles are known to possess the kineticenergy 5.3 MeV& practicallyall daughternuclei are formed directlyin
The halflifeofthis alpha emitter is 138 days.
(a) What mass of^'®Po isneeded topower athermoelectric cell of 1 W output if the efficiency of energy conservation is 8 percent ? (b) What would be the power output after 1 year ? Ans. [88.4 mg, 0.16 watt]
the ground state. Ans. [s 1.6 MJ]
4107 To investigate the Pdecay of ^^Mg radionuclide a counter was activated atmoment r=0. Itregistered Nj Pparticles by a moment = 2.0 s and a moment = 3/,, the number of registered Pparticles was 2.66 time greater. Find the mean life
4101
decays with a halflife of4.51x10^yrs, thedecay time of the given nuclei. [TakeIn (0.88)=  0.125]
series eventually ending at
which is stable. A rock sample
analysis shoes that the ratio of the numbers of atoms of ^®®Pb
to^•'^U is0.0058. Assuming that allthe^'^Pb has been produced by the decay of and that all other halflives in the chain are negligible. Calculate the age ofthe rock sample. Ans. [38 X 10'' yrs]
Ans. [t = 16 s]
4108 Find the decayconstant and the mean life time of ^^Co radionuclide if its activity is known to decrease 4% per hour. The decay product is nonradioactive. Ans. [X  1.1 X 10"^ s"'.]
Nuclear Physics and Radioactivity ;
:i86
4109 Taking into account the motion of the nucleus of the nucleus of a hydrogen atom, find the expressions for the electron's binding energy in the ground state. How much (in percent) do the binding energy obtained without taking into
4113 Find the greatest possible angle through which a deuteron is scattered as result ofelastic collision with an initially
stationary proton ? Take w, & Wj as masses of a proton & a deuterium.
account the motion ofthe nucleus differ from the more accurate
corresponding valueofthisquantity. Given
ut
Ans. [S , = sin '—^ = 30°]J LPmax =0.00055, where
m and Mare the masses of an electron and a proton. Ans. [
. , 0.055%. Here h =
and n =
]
4114 The element Curium
has a mean life of
10'^ seconds. Itsprimary decay modes arespontaneous fission and adecay modes are spontaneous fission and adecay, the former with a probability of 8% and the latter with a probability
4110 An a  particle with kinetic energy
= 7.0 MeV is
of92%. Each fission releases 200 MeV ofenergy. The masses
scattered elastically byaninitially stationary ^Li nucleus. Find
involved in adecay are as follows : (1 u = 931 MeV/c^). Calculate thepower output from asample of10^° Cm atoms.
the kinetic energy ofthe recoil nucleus if the angle ofdivergence of the two particles is 0 = 60°. Take masses of a particle and lithium as m and Mrespectively. Ans. [r =
[1 +(M m)^ / 4ffjA/cos^ 0]
Ans. [3.32 x 10"^ Js"']
4115 Nuclei of radioactive element.^ are being produced at a constant rate a. The element has a decay constant X. At time
= 6.0 MeV]
/ = 0, thereAq nucleiofthe element. 4111 A neutron collides elastically with an initially stationary deuteron. Find the fraction of the kinetic energy lost by the neutron. Take masses of neutron and deuteron as m and M
respectively. (i) in a headoncollision.(ii) in scattering at right angles. Ans. [(a) T) = 4 Mm {m + M)^ = 0.89, (b)
(a) Calculate the number N ofnuclei ofA at time t.
(b) If a = 2NqX, calculate the number of nuclei ofA after one halflife of^ and also the limiting value ofiVas / ^ oo.
Ans. [(a) Af =^ [a  («  XN,)
(b) 2%]]
2m
{m+M)
4116 The mean path length ofaparticles in air under standard
conditions is defined bythe formula i? = 0.98 x 10"^' 4112 Find the energy of the reaction (a , p) if the kinetic energy ofthe incoming aparticle is =4.0 MeV & the proton outgoing at an angle 0 = 60° to the motion direction of the aparticle has a kinetic energy = 2.09 MeV. Ans. [Q = {1 + Tip T; (1  nj T^2
cos 9 =  1.2 MeV,
cm,
where Vq (cm/s)isthe initialvelocity ofan aparticle.Usingthis formula, find for an aparticle with initial kinetic energy7.0 MeV. (a) Its mean path length. (b) The average number of ion pairs formed by the given aparticle over the whole path R as well as over its first half. Assuming the ion pair formation energy to be equal to 34 eV.
wheren
Ans. [(a) 6.1 cm (b) 2.1 x 10^ and 0.77 x IQ^]
Geometrical Optics FEW WORDS FOR STUDENTS
Image
The topic of ray optics or geometrical optics is concerned with the analysis ofpropagation of light in a medium by considering it as the propagation of a 'Ray\ A ray defines
Mirror
the path,.along with lightpropagates and it can be assumed as an infinitesimally thin pencil oflight moving in a specific
direction at a given point in space. In this topic ofray optics we do not bother about the nature oflight and we ignore the wave and photon character oflight. Here wefocus on how light behaves on a large scale.
CHAPTER CONTENTS
5.1
Understanding a Light Ray andLight Beams
5.12
Total Internal Reflection
5.2
Reflection ofLight
5.13
Refraction by a Prism
5.3
Understanding Object and Image in Geometrical
5.14
Refraction by a Thin Lenses
Optics
5.15
Analysis ofImage Formation by Thin Lenses
5.4
Reflection and Imageformation by a Plane Mirror
5.16
5.5
Field of Viewfor Imageformed by a Plane Mirror
5.6
Characteristics ofImageformed by a Plane Mirror
5.7
Understanding Shadow Formation
5.8
Spherical Mirrors
5.9
Analysis ofImageformation bySpherical Mirrors
5.10
Refraction ofLight
5.11
Refraction ofLight by Spherical Surfaces
Optical Power ofa Thin Lens and a Spherical Mirror
5.17
Lenses and Mirrors submerged in a Transparent Medium
5.18
5.19
Displacement MethodExperiment to measure focal length ofa Convex Lens ' Dispersion ofLight
5.20
Optical Aberrations in Lenses and Mirrors
5.21
Optical Instruments
COVER APPLICATION
Figure(a)
Figure(b)
Figure(a) shows the experimental setup of refraction of light through a trihedral prism and understanding of minimum deviation of light and figure(b) shows the ray diagram of light refraction through the prism.
Geometrical Optics,
188'
In this chapter of light, we will involve only geometric considerations for light propagation and all laws are formulated using the concepts of geometry thats why, ray optics is also called ^Geometrical Optics\ The basic concept we use is that a light ray travels along a welldefined path. Ray optics in reality is an approximation but it is very important when character and effects of light are studied at macroscopic level. Topic of ray optics is very important in understanding image formation by mirrors, lenses and it helps in studying the working ofdifferent types of optical instruments. Ray Optics is a very important branch of physics but many aspects of light cannot be explained on the basis of ray optics likeinterference, diffraction and polarizationoflight which require microscopic level understanding of wave theory oflight and such concepts and theories we will study in next chapter of ''WaveOptics'.
5.1.1 DifferentTypesofLightRays
When light rays incident on an optical device like a spherical mirror, a spherical surface, a lens or an optical instrument then in reference to the optical device light rays are categorized in two ways paraxial rays and marginal rays. Lets discuss these in detail.
Paraxial Rays: These are the light ray^ which incident on an optical device very close to its principal axis and make very small angle with it or which are nearly parallel to principal axis of the device. In case ofsome lenses and mirrors paraxial rays are shown in figure5.3. Most ofthe analysis ofimage formation in geometrical optics, we will study in this chapter are restricted to paraxial rays only.
5.1 Understanding a Light Ray and Light Beams In previous grades you have studied that light travels in a straight line that's what we call '^LawofRectilinear Propagation ofLight'. This is the concept based on which we say that an object placed in the path oflight coming from a source produces a sharp shadow on a flat surfece or a screen. This law ofrectilinear propagation is considered valid only at macroscopic level or large scale. Light does not travel in a straight line when the size ofobject or obstacle is ofthe order of wavelength oflight. Such concepts we will study in microscopic analysis oflight under wave optics and not considered here.
1
h^,
0, 02 are very small
As already explained that a light ray can be imagined as a very thin pencil oflight travelling in a specific direction. It is represented by drawing a straight line with an arrow in the direction oflight propagation as shown in figure5.1. A bundle of several light rays is called a light beam which is shown in
figure5.2. For analysis of ray optics it is very important to understand different types of light rays and light beams in reference to an optical device (generally a mirror or a lens) or any optical instrument on which the light rays incident as the concept of image formation by any optical device involves a light beam through which light rays incident on the optical device.
hy h^, 63, 04are verysmall Figure 5.3 Figure 5.1
Figure 5.2
Always remember that light rays as such do not exist in reality and by any method light rays cannot be isolated experimentally from a light beam,these exist only in theoreticalunderstanding.
Marginal Rays: These are light rays which makes large angles to the principal axis of the optical device falling on it. Figure5.4 shows some marginal rays in different cases of optical devices. Images formed by marginal rays are blurred and distorted so
•Geometrical Optics 189
these rays are not generally used in image formation while propagation and all light rays appear to converge at apoint in analyzing the cases under geometrical optics. space which is called the point of convergence of the light beam. Agenera! convergent beam oflight is shown in tigure5.6.
Figure 5.6
Generally such abeam is incident on an object (or optical device) which is placed between the light beam and its point of convergence as shown in figure5.7(a). Ifthe object (oroptical
device) is placed beyond the point of convergence then the incident beam will be diverging not converging as shown in figure5.7(b)
hy hy
9,, 0^ and 0^ are large Figure 5.4
Converging incident rays falling on object (a)
5.1.2 Different Typesof Light Beams
From a source oflightalways several rays areemitted in form of
a beam. A single light ray cannot be considered as only ray emitted from a lightsource. Depending upon the behaviour of
light propagation, light be.jm is categorized in three ways Convergent Beam, Divergent Beam andParallel Beam oflight. Lets discuss these in detail.
Diverging incident rays falling on object
Divergent Beam of Light: This isa light beam in which the
(bj
diameter of light beam increases in the direction of light propagation andall light rays appear tobecoming from a point opposite to the direction of propagation of light. A general
Figure 5.7
divergent beam oflight isshown infigure5.5(a). From a luminous pointsourceof light diverging beam of light is emitted in its surrounding in all direction as shown in figure5.5(b)
Parallel Beam ofLight: This isa light beam in which all rays constituting the beam move parallel to each other and the
diameter ofthe beam remains same throughout the propagation oflight. Aparallel beam oflight isshown in figure5.8. i\
/•
I, Figure 5.8
5.2 Reflection of Light (a)
(b)
Figure 5.5
ConvergentBeamof Light: Thisisa lightbeam inwhich the diameter of light beam decreases in the direction of light
Whenever a lightrayincident on a boundary oftwomedia then
apartoflight isbounced back into thesame medium and a part goes into the other medium as shown in figure5.9. This phenomenon ofbouncing oflightenergy intothe medium from
Geometric^
;i90
(
which light incident on the boundary is called 'Reflection of gets reflected in alldirections from the point sothe light spot on the surface can be seen from all directions. Due to diffused
Light'.
reflection only when we puta parallel light beam ofa torch ona wall, thespot oflight formed can be seen from any location in theroom as lightrays in the beam arereflected in alldirections randomly in theroom from thesurface where theincident beam
N
Medium1 Medium2
falls.
Figure 5.9
Transmission of light energyinto the other medium is called 'Refraction ofLight'.Inthissection wewill study reflection of light indetail then later we will cover refraction oflight. Reflection oflightfrom a surface or a medium boundary isclassified intwo ways. These are 'Regular Reflection' or 'Specular
Surface with
irregularities Diffused reflection
Figure 5.11
Reflection' and 'DiffusedReflection'. 5.2.1 Regular or Specular Reflection
Whena lightbeam incident on a perfect planesurface then the reflection is called Regular or Specular Reflection. In such a case the reflected light beam has high intensity only in one
Rough surface
direction which is the direction of propagation of the reflected beam as shown in figurerS. 10. Figure 5.12
5.2.3 How we see an object in our surrounding
Perfectly
plane surface
When from any luminoussourceoflight, light raysincidenton an object, due to the roughness of the surface of object these raysare reflected from the object surface in diffused mannerin all directions so wherever an observer is situated, she will be
able to see the face of object in front of her eye as shown in Regular reflection
figure5.13
Figure 5.10
Angle ofDeviation:'Infigure5.10 thetotal angle bywhich the light ray is'rotated which is shown as 5 is called angle of deviation, the angle between the direction of propagation of initial ray and final ray.
5.2.2 Irregular or Diffused Reflection
When a light beam incident on a surface which is rough or havingirregularities thenthe lightrays in the incident beam of light will be reflected in irregular behaviour as shown in figure5.11. Eachlightrayofthe beam isreflected from"the local point on the surface on which it incidents and different light rays are reflected in random directions depending upon the irregularities onthe surface. Thefigure5.11 isa highly enlarged viewofa rough surface, if we lookonto it normallyit lookslike figure5.12 whereon a point ifa narrowlight beam incident,it
Figure 5.13
For understanding of observer's view it is very important to understand how our eyes perceive the shape, size and colour
^Geometrical Optics 191 •
ofan object. Everyobject in our surrounding which is not point Second Law ofReflection : The angle ofreflection is equal to
sized (these are called extended objects) can be considered asa the angle ofincidence soaccording tothis law we have combination of several points on its surface and from each
point on the surface ofobject, light rays falling on it will be irregularly (diffused) reflected in all directions so from any side ofobject onwhich anylightis falling, it is visible toobserver's
eyes inwhich irregularly(difl^ised) reflected light rays from the surface are incident as shown infigure5.13.
...(5.1)
I =r
5.2,5 VectorAnalysis of Laws ofReflection
Laws ofreflection can be analyzed with the help ofvector algebra also by considering unit vectors in the direction of incident
Thelight rays which areirregularly reflected from thesurface of rays, reflected rays and normal to the boundary. Keeping above
object carry information ofcolor and its brightness and from all points ofsurface and boundaries ofthe object these rays fall
two laws of reflection we relate these unit vectors with the
anglesof incidence and angleof reflection.
into observer's eye simultaneously due to which observer is able tosee thewhole object, itsspecific size, shape andcolour. In the figure5.16, reflection ofa light ray incident on aplane surface is shown. If we consider i, r and n as unit vectors
along thedirection ofincident ray, reflected rayand normal to thesurface asshown then first we can write components of i Laws of Reflection governs the way light is reflectedfrom the and r intermsoftheunit vectors along thenormal and along surface of a boundary of two different media/ There are two the surface. Here we are considering i as a unit vector along 5.2.4 Laws of Reflection
reflection laws which explains how light is reflected from a
the surface so these are given as
surface.
When a light ray isincident on a surface then the angle itmakes with thenormal tothesurface iscalled 'angleofincidence' and after reflection, theangle which thereflected raymakes with the normal is called 'angle ofreflection'. Irithe figure5.14 these angles are represented byangle i and angle r respectively.
i = (sin 0)i  (cos 0)h
...(5.2)
r = (sin 0) t + (cos 0) n
...(5.3)
['NrHnNhl=l
mediuml
(lsin0)
(lcos0)
mediumI!
Figure 5.14
First LawofReflection: Inthereflection process theincident ray, thereflected rayandthenormal at thepointof incidence lie insame plane. This plane iscalled 'PlaneofIncidence' asshown
(1 cos 0)
(1 sin 0) Figure 5.16
Now from equations(5.2) and (5.3), subtracting these we get
in figure5.15.
r=i+ (2cos 0)«
... (5.4)
From the dot product of i and h we have Plane of Incidence
i. « =  cos 0
...(5.5)
From above equations we get r = i 2{i .h)h
•
...(5.6)
Above equation(5.6) is called the equation of lawsofreflection which account for both the laws as vector form includes the
plane ofincidence as well as equal angle ofincidence and angle Figure 5.15
ofreflection as shown in figure5.16.
^192'
Geometrical Optics
y ' _ _
and finally after reflection through the second plane mirror placed along xz plane theunit vector oftherayisgiven as
# Illustrative Example 5.1
Arayoflight is incident on the(yz) plane mirror along a unit
vector e, = +^7+^^. Findtheunitvectoralongthe reflected ray.
1 ^ F'
e, =.
41
1 \ r 7ry+?r^
43' 43'
# Illustrative Example 5.3
Solution
A ray of light is incident on a plane mirror along a vector
The normal unit vector to the (y  z) plane mirror is along
i\jk. Thenormal onincidence point isalong 1+ j. Find a
X direction so we have
unit vector along the reflected ray. h = i Solution
xii)
Ascomponent of incident rayalongthe normal getsreversed while the component along the surface remains unchanged. Thus the component of incident ray vector A= i+jk Plane Mirror (yz plane) Figure 5.17
parallel tonormal, i.e., i+j gets reversed while perpendicular
Then the unit vectoralong reflected ray is given as ^2 and the
to it,.i.e. alongthe surface, k remains unchanged. Thus,the reflected ray can be written as
relation between Cj and e, is given as
^2 ~ ^1
R=l]~k
•n)h
&=( ' ; _ 2(4)i 'S S' S' 1 : e, =
1
The unit vector along the reflected ray is given as
1
,
7r'
R
'~R~
ijk
43
Or we can directly say that the component of the incident ray
along the normal gets reverses i.e. along unit vector i which
r =
0+J + k)
directly gives the final result.
.5.3 Understanding Object and Image in Geometrical
# Illustrative Example 5.2
Optics ^
Two plane mirrors arecombined toeach other as such oneis in (yz) plane and other is in (x z) plane.A ray of light along In analysis ofvarious situations andcases ofgeometrical optics, it is very important to have a clear understanding of the terms is incident on the first mirror. Find
'Object^ and ''Image'' in the given situation otherwise lot of
the unit vector in the direction of emergence ray after
confusion arise and even simple problems seem very complex. First of all students must keep one thing in mind that these terms Object' and Image' in geometrical optics are considered
vector
S
43
+
43 '
successive reflections through these mirrors.
with respect to a specific optical device which is generally a Solution
Just like the case of previous problem here we can see that
mirror or a lens or it can be any type of optical instrument or human eye also. Lets first define an object and an image.
after reflection from the mirror alongyz plane the component of
the incident ray along x direction gets reverses so the unit vector of the ray after reflection from first mirror is given as \
I
.
5.3.1 Object in Geometrical Optics
For any given optical device, an object is considered as the intersection point ofall the incident rays falling on the optical
"Geon^trical Optics 193
device. For existence ofobject for adevice it is not necessary that these incident rays really intersect. Objects can be point sized or extended, as shown in figure5.12 and 5.13 ifobject is extended then every point on surface ofobject isconsidered as apoint object and combination ofall these point object is taken as the extended object.
Virtual Object
There are three wa)^ in which object is classified in geometrical optics depending on the type oflight rays incident on the optical device. Lets understand these cases ofvarious incident rays
Optical Device Figure 5.19
on a device for a point object.
CaseIll: ParaUel Incident Rays on anOptical Device CaseI: Diverging Incident Rays on anOptical Device
When light rays incident on an optical device are parallel then
When light rays incident on an optical device are in diverging in this case we consider object is located at infinity as parallel manner then their intersection point will be in the direction
rays can only be considered intersecting at infinity. So when
opposite to the direction of propagation of light as shown in
object is located at infinity, we do"not talk about its nature whether realor virtual. Figure5.20 below shows the case when
figure5.18. This point is always regarded as a"Real Object' for the optical device irrespective ofwhether light rays are really parallel incident rays incident on an optical device. intersecting on this point (actually coming from this point) or not. The type of object (in this case a real object) is defined
only by the type of light rays in the vicinity ofthe optical device and not from the actual source from where light rays are coming.
Object is at infinity Optical Device
Figure 5.20
Real Object
5.3.2 Image in Geometrical Optics Optical Device
For anygiven optical device, an image is considered as the Figure 5.18
Note: Most ofthe irregularly reflected light rays from objects
intersection point ofall the reflected or refracted rays from the
device depending whether the device is areflecting or refracting
one. Based on the type ofreflected or refracted rays coming in our surrounding fall inour eye indiverging manner asshown from the device, image can also be classified in three ways in in figure5.12 so for our eye (or for an observer's eye) what we geometrical optics in reference to the device producing the see inour surrounding can be taken asa "Real Object'. image. CaseII: Converging Incident Rays on anOptical Device
CaseI: Diverging Reflected or Refracted Rays from the
When light ra>^ incident on an optical device are in converging
Optical Device
manner then their intersection point will lie in the direction of When light rays coming from the optical device are in diverging propagation of light behind the optical device as shown in manner then their intersection point will lie behindthe device figure5.19. These lightrays cannot intersect reallyin such a opposite to the direction of propagation of light as shown in case and in this case this point O is regarded as a "Virtual figure5.21. This point isalways regarded as a ' VirtualImage'
Object' for the optical device. In this case also the type of for this optical device. Always remember that for diverging light object (virtual object in this case) isdefined only by the type of rays it is not possible to reallymeet or intersectin the direction light rays in the vicinityof the optical deviceand not fromthe of light propagation so such reflected or refracted rays are actual source from where light rays are coming.
considered to produce a virtual image.
Geometricai Optics!
;i94
Diverging Reflected or
Refracted rays
Ifwe look atthe figure5.24, here for optical deviee1,1 isa 'Real Image" as reflected or refracted rays from this device are converging but before actually intersecting at I light rays incident on another optical device2 so for this second device1 will actlike a' Virtual Object", as shown infigure5.24.
•e?:.
I
^^30°J
=>30° (clockwise)
Figure 5.62
[12]
Geometrical Optics j
l20;e
(v)
A point source of light B is placed at a distance L in
front ofthe centre ofa mirror ofwidth d hung vertically on a wall. A man walks in front ofthe mirror along a line parallel to the mirror at a distance 2L from it as shown in figure5.63. The greatest distance over which he can see the image of the light
(ix)
A plane mirror placed along xz plane is moving with
velocity 3f +5y4/fc. A point object in front of the plane mirror ismoving with velocity 2z 4y+4^. Find velocity of image.
[2/ +14j+4i ]
source in the mirror is:
(x) Two plane mirrors are inclined to each other as shown in figure5.66. Arayafler the three successive reflection falls
j
on the mirror Mj and finally retraces its path. Calculate the angle between the two plane mirror.
11
Figure 5.63 [3d]
(vi) In a 3D coordinate systema plane mirror is placed parallel to XYplane and above the mirror a point object is moving at
velocity Vq =5i 3JIlk m/s.Ifmirrorisalsomovingparallel toitselfat velocity = 2z"+y3yt, find thevelocity ofimage
Figure 5.66 [15°]
produced in mirror.
[5/3j +5^ m/s]
5.8 Spherical Mirrors
(vii) Two planemirrorsA/, and are inclined at angle0 as shown in figure5.64. Aray oflight 1, which is parallelto Mj
A spherical mirroris a part ofhollow glassspherepolished on
strikes
and after two reflections, the ray 2 becomes parallel
to Mj. Find the angle9.
one ofits surface. Ifinner surface ofsphere is polished then its outer surface becomes reflecting and it is called a convex mirror as shown in figure5.67. Reflecting face //'
\\
Policed face
(a)
Figure 5.64
Figure 5.67
(viii) An object moves with 5 m/s towards right while the
Ifthe outer surface ofthe hollow spherical part is polished then its inner surface becomes reflecting and it is called a concave mirror as shown in figure5.68.
W]
mirror shown moves with 1 m/s towards the left as shown in
Reflecting face
figure5.65. Find the velocity ofimage. object »
>
5 m/s
1 m/s
j: mirror
(b) Polished face
Figure 5.65 [7 m/s]
(a)
Figure 5.68
jGeometrical_Optics
207
5.8.1 Standard terms related to Spherical Mirrors For concave or convex mirrors there are some specific terms
and their definition associated which are useful in analyzing image formation by these mirrors so first we will discuss all these terms one by one. Focus of concave mirror
(i) Pole or Optic Center of Mirror : It is the center of the
(a)
physical sphericalmirror as denotedbypointP in figure5.69. (ii) Center of Curvature of Mirror : It is the center of the
spherical shellof whichmirroris a part. It isdenoted bypoint C in figure5.69 and 5.70
(lii) Principal Axis of Mirror: It is the line joining the optic center of mirror and its center of curvature as shown in
P
'
*
\
figure5.69. (iv) Radius of Curvature of Mirror ; It is the radius of the spherical shell ofwhich the mirror is a part which is the distance OC and it is also denoted by'/?' as shown in figure5.69.
(v) Focal point of Mirror : It is the mid point ofline joining the optic center and center of curvature and denoted by 'F". This is a point where it is considered that all paraxial light rays falling on the mirror which are parallel to principal axis will converge in case of concave mirror as shown in figure5.70(a) and appear to diverge from this point in case ofconvex mirror as shown in figure5.70(b). It is also called 'Principal Focus' of the spherical mirror. (vi) Aperture of the Mirror : It is the diameter of the actual mirror cross section through which light rays incident will fall on the mirror surface. This is shown as 'tf in figure5.67(b) and 5.68(b) and asAB in figure5.69(a) and 5.69(b)
Focus of convex mirror
(b)
Figure 5.70
5.8.2 Focal Length of a Spherical Mirror
It is the distanceofPrincipalfocus of mirror to the OpticsCenter of the Mirror. Generally concave mirrors are also referred as
'Converging Mirrors'' because these converge all parallel rays incident on it. This convergenceoccursat focalpoint if incident rays are parallel to the principal axis of mirror as shown in figure5.70(a).In this manner all convex mirrors are also referred as 'Diverging Mirrors' because these diverge all parallel rays incident on it. If incident rays fell on a convex mirror parallel to the principal axis then all these rays diverge in such a manner that these appear to be diverging from the focal point of the mirror as shown in figure5.70(b). The situation slightly differswhen incidentrays are not parallel to principal axis. In this case also concave mirror converges these rays and convex mirror diverges these rays but the point ofconvergence and divergence like in a plane normal to principal axis passing through the focus of mirror as shown in figure5.71(a)and 5.71(b).
Principal axis
(a)
3"^ %
1^^ 1
c
% g
d
B
(b) Figure 5.69
Focal plane
Geometrical Optics !
208.
willnot be a sharp point, it will get blurred.At the end ofchapter we will discuss on the defect in image formation ifrays are not paraxial under the topicof sphericalaberration.
Focal planci
5.8.4 Standard Reflected Light Rays for Image Formation by Spherical Mirrors
Forunderstanding ofimageformation bya spherical mirror, we need to consider some standard light rays incident on the mirror and their corresponding reflected rays. If any two of these rays incidenton a spherical mirror and their correspondingreflected rays are used then image can be obtained by finding the
(b)
Figure 5.71
5.8.3 Image Formation by a Spherical Mirror using Paraxial Rays
intersection point of the these two reflected rays as we have already discussed that all the reflected rays intersect at the samepoint(image)if all incidentraysare paraxial. Letsdiscuss four standard rays used in image formation by ray diagram for spherical mirrors.
It is observed that for a specific point object placed in front of a spherical mirror all the light rays incident on the mirror are reflected such that these all reflected rays converge or appear
to diverge only fromone point which is consideredas 'Image^of the'Object" produced bythe mirror.Figure5.72(a) and (b)shows
images /, and formed bythemirrors A/, and Here wecan see that image /, isproduced isa 'RealImage" because reflected ray are converging and image produced is a' VirtualImage" because reflected rays are diverging.
Ray1: Incident Ray parallel to PrincipalAxis
Figure5.73(a) and (b) shows a rays incident on concaveand convex mirrors which is parallel to principal axis. Such a ray after reflectionpasses through the principal focus of the mirror as shown. In case ofconcave mirror it actually passes through the focus and in convex mirror it gets reflected such that it appear to pass through its focal point In virtual space behind the mirror.
Object
Real imaae
Real focus
(a)
Object
Virtual
image
'V
Virtual focus
Figure 5.72
Note : Above cases which we discussed for image formation by a spherical mirror is valid only when we consider incidence ofonly paraxial rays from object to mirror. Ifincident rays are not paraxial then all the reflected rays will not converge or appear to diverge from one single point. In this chapter we will mainly consider the image formation by paraxial rays only. If rays considered are not paraxial then for a point object image formed
(b) Figure 5.73
Ray2: Incident Ray passing through Center ofCurvature
Figure5.74(a) and (b) shows a rays incident on concave and convex mirrors which passing through center of curvature. Such a ray falls on mirror normal to its reflecting surface and gets reflected as it is and retraces the path of incident ray.
iGeornetrical Optics _
^
20£j
Ray4: IncidentRayIncidentonPoleofthe Mirror
Figure5.76(a) and (b) shows a rayincident on concave and convex mirrors onitspole. Atthis point themirror normal isthe principal axis only so the light ray isreflected asifitisreflected
,
c\
F
from aplane mirror atthe same angle from principal axis at which it incidents asshown infigure.
Real focus
Real focus
(a)
Virtual focus
(a) Figure 5.74 Virtual focus
Ray3 : Incident Raypassing through Focus
(b)
Figure5.75(a) and (b) shows a rays incident on concave and convex mirrors which passing through focus ofthemirror. Such
Figure 5.76
arayafter reflection gets parallel tothe principal axis ofmirror
5.8.5 Relation in focal length and Radius of Curvature of a
as shown. In case ofconcave mirror lightactually passes through ^SphericalMirror focus and then afterreflection becomes parallel to the principal axis butincase ofconvex mirror itincident onthemirror insuch Figure5.77 shows alight ray which is parallel to principal axis a waythat it appearto pass through virtual focus ofthe mirror
ofthe mirror shown and itincident atapointyf ofthe mirror atan
and after reflection itbecomes parallel to the principal axis as
angle 0to the normal (dotted lineAC) atthis point which passes through the center ofcurvature ofthe mirror. According to laws ofreflection it will get reflected at the same angle 8and pass through apoint Bon principal axis as shown in figure. Now in A
shown in figure.
ABC we can use AC^R^lBCcos 0 R
BC =
2cos0
Real focus
\
P
N
P
Figure 5.77 s,
1
1
F
—
1
—
C
Thus focal length ofmirror can begiven here as distance PBso we have
Virtual focus
PB = R~BC
(b)
Figure 5.75
PB^R
R
2cos0
...(5.9)
:
"
_•
Iftheincident light ray isparaxial then 0 will besmall so we can usecos 0~ 1soPB=f=R/2. Equation(5.9) isgives thedistance PB where parallel marginal rays after reflection will intersect on
'Geome^ical Opticsj
Casen; Objectislocated beyond CenterofCurvature
Figure5.79 shows this situation in which object is asmall candle andwe find theimage oftipofthecandle byconsidering ray1
principal axis. Distance PB can be considered as focal length and ray2 as explained inarticle5.8.4. With this ray diagram we onlywhen incident rays areparaxial.
canconclude thattheimage produced for thislocation ofobject
5.8.6 Image formationby ConcaveMirrors
(placed beyond Con principal axis) isreal, located between F and C, inverted (on other side ofprincipal axis) and smaller in size than that ofobject.
Whenever an object is specified for a spherical mirror then
using any two of the four incident rays and corresponding reflected rays mentioned inarticle5.8.4 we can find theimage
by using ray diagram. There are some positions near to principal axis indifferent regions infront ofthemirror where ifanobject iskept, using raydiagram wecan getsome information about theimage produced. This information isvery helpftil inrough analysis of image formation. For both concave and convex mirrors we aregoing to discuss different cases for position of
Figure 5.79
object in front of the mirror and its corresponding image Casem: Objectislocated at Center ofCurvature produced. First we will take up the cases for concave mirror in Figure5.80 shows this situation inwhich object (candle) located which five possibilities are there for placement ofa real object at C and wefindthe imageof the candle byconsidering ray1 in front ofthe concave mirror.
andray4 asexplained inarticle5.8.4. With this raydiagram we
CaseI: Objectislocatedatinfinity
can see and concludethat the imageproducedfor this location
ofobject (placed at C on principal axis) is real, located at C, We have already discussed the case when object is located at inverted (on other side ofprincipal axis) and ofsame size asthat infinity, incident rays felling on the mirror will be parallel and for of object.
parallel incident rays there are two possibilities. One is when all
¥
incident rays are parallel toprincipal axis when image produced 0
isrealand located at focus ofthemirror asshown infigure5.78(a)
Object
and other possibility iswhen these are atsome angle toprincipal axiswhen image produced is realandlocated in focal planeas shown in figure5.78(b). In both of these cases we have considered the image produced is highly diminished in size, realand inverted (produced onthe other sideofprincipal axis
C
Image
Figure 5.80
as that of object).
CaseIV: Objectis located at between Focus and Center of Curvature
Figure5.81 shows this situation in which object (candle) is locatedbetweenF and C and we find the image of the candle by
considering ray1 andray3 asexplained in article5.8.4. With this ray diagram we can see and conclude that the image (a)
produced for this location ofobject (placed between F and C onprincipal axis) isreal, located beyond C, inverted (on other side ofprincipal axis) andenlarged compared tothat ofobject. M
(b) Figure 5.78
Figure 5.81
[Geometrica{ Optics
^^
211:
CaseV: Object is locatedat Focus
CaseI: Object islocated at infinity Figure5.82 shows this situation in which object (candle) is When light rays from distant object fall on a convex mirror, located atfocal point ofthe mirror and we find the image ofthe these rays diverge after reflection insuch awaythatfor incident candlebyconsidering rayl and ray4 asexplained inaiticle5.8.4. rays parallel toprincipal axis image isobtained atfocal point of With this raydiagram we can see that thereflected rays are mirror as shown in figure5.84(a) and fer incidentrays non parallel parallelandwillproduce image at infinity. Sowecanconclude with this diagram thatimageproduced will beatinfinity, inverted (on theother sideofprincipal axis) andhighly enlarged.
toprincipal axis image is obtained in focal planeas shown in figure5.84(b). Theimage produced will bevirtual, diminished
and erected (produced on the same side ofprincipal axis where object is located).
Figure5.82
CaseVI: Object is located between Focus and Pole
Figure5.83 shows this situation in which object (candle) is located between Oand F and we find the image ofthecandle by considering ray1,ray2andray4 as explained in articIe5.8.4.
With this raydiagram we can see that allthethree reflected rays fi^om mirror are diverging inaway that these appear tobe coming from thepoint/behind themirror from a virtual image. So for thislocation ofobject (placed at F onprincipal axis) image is
(a)
produced behindthe mirror, virtual,erected (on the sameside ofprincipal axis) and enlarged. Focal plane (b)
Figure 5.84
CaseII: Object is placed anywhere in front ofmirror
Figure5.85 shows a situation in which the object (candle) is placed infront ofthe convex mirror and tofind image using ray Figure 5.83
diagram we consider rayl, ray2 and ray4 as mentioned in Above five cases are general cases which explains the relative article5.8.4. We canseein thisfigure thattheimage is formed position, nature andorientation ofimage produced bya concave byback extension ofthereflected rays corresponding tothese mirror for a real object placed close to itsprincipal axisusing incident rays from theobject andtheimage is produced behind paraxial rays. While solving different questions above cases themirror between F'and O, virtual, diminished and erected (on gives an idea about the estimation of image produced upto thesame sideofprincipal axis as that ofobject). some extent before going for the exact process of image formation.
5.8.7 Image formation by ConvexMirrors
Unlike to the different cases of imageformation bya concave
Image f
mirror for different positions of object in front of it, in case of
convex mirror there areonlytwo possibilities ofimage formation for a real object placed in front ofit. Figure 5.85
Geometricai Optics ^
:212
5.8.8 How an observer sees image of an extended object in a spherical mirror
Figure5.86(a) shows an object (candle) placed in front of a concave mirror placed between its pole and focus. This we have already discussed in caseV of article5.8.6. From the tip and bottom points of the objectsAB the light rays are incident on the mirror gets reflected and falls into the eye of observer. Here observer simultaneously sees the image oftop and bottom points of objectas A' and B' and simultaneouslyfor all points in between which constitutes the full image of candle yl'5'. For objectsplaced close to principal axis we consider all rays incident
on mirror are paraxial so we can find the image of the tip ofthe object in any case and drop the normal from this point to the principalaxistogetthe fullimagein all suchcases. Figure5.86(b) shows the similar case for an observer seeing the image produced by a convex mirror.
Field of view to
see the image (b)
Figure 5.87
5.8.9 How image produced by a spherical mirror can be obtained on a screen
(a)
A screen is generally an opaque and rough white surface. Any point of which if light rays fall are diffused reflected in all direction from the point of screen. If a light beam falls on the screen than it makes a light spot on screen which can be seen from any direction in front of it as everypoint in the spot light rays are diffused reflected by screen in all directions as shown in figure5.88. In this figure the light spot formed due to the
light beam can be seen by all the observers Oj,
and O3
shown who are looking at the spot from different directions.
(b) Figure 5.86
Ifobject is point sized and placed on the principal axis then its image will also be produced on the principal axis as shown in figure5.87(a) and it can also be seen by the observer as shown. Figure5.87(b) shows the field of view by the shaded region bounded by the reflected rays from the edges ofthe aperture of the mirror from which the image can be seen as reflected rays exist in this region only so to view this image observer eye must lie in this region.
03 Figure 5.88
Figure5.89 shows a case when image formed by a concave mirror for an object (point light source) is obtained on a screen.
IGeometrical Optics 213
We can see in figure5.89(a) when screen is placed exactly at the location ofimage produced then all thelight rays reflected
from the mirror will be meeting at point/on screen and abright and sharp image can be obtained on the screen. Ifwe displace the screen slightly toward the mirror, sharp image will not be
obtained on screen and ablurred spot is produced due to light
Blurred image
rays falling onthescreen before meeting atpoint/as shown in
figure5.89(b). Ifwe displace the screen slightly behind the image then also we can see that a blurred spot will be formed on
(b)
screen rather than sharp image asshown infigure5.89(b).
Blurred image
(c)
Sharp image on screen
Figure 5.90
(a)
With the above analysis it is clear that sharp image can be obtained on a screen only if screen is placed exactly at the location ofimage. Anther fact about images obtained onscreen is that onlyreal images can be obtainedon screen becausereal
images are produced byconverging rays. Diverging rays falling Image gets blurred (b)
ona screen can only produce brightness onscreen asnosharp boundaries can bemade bydiverging rays onany surface. 5.8.10 Sign Convention
For agiven spherical mirror, object and its corresponding image may lieanywhere either in front ofthe mirror ifimage isreal or behind the mirror ifimage is virtual. For mathematical analysis ofimage formation by a spherical mirror, we associate a sign convention with thegiven situation ofmirror and inanalysis of image formation all the distances ofobject, image, focal length
Image gets blurred
orradius ofcurvature ofmirror, wemeasure, willbeconsidered (c)
Figure 5.89
Figure5.90 shows the formation ofimage ofan extended object (candle) on ascreen. Here in figure5.90(a) the screen is placed exactly at the location ofimage in which thesharp image of candle is obtained on screen which can be seen from any direction ifwelook atscreen. But ifscreen isslightly displaced along principal axis inanydirection, theimage will get blurred as shownin figure5.90(b) and5.90(c).
with proper '+' or
signs according tothe sign convention
used.
There are several sign conventions invented inpast tobe used in geometrical optics for different cases of reflection and
refraction. We are discussing here two sign conventions both of which are very popular for analyzing image formation in differentcases of reflection and refraction. In this bookwe will be using the first convention however ifstudents wish then for
each case studied they can use second convention and.verify that the results obtained are same. It is most important to understand and follow a sign convention properly.
FirstConvention: Incident RayReference Sign Convention
This sign convention is also called 'New Cartesian Sign (a)
Convention'. Ifwelook atthesituation shown infrgure5.91(a). It shows a concave mirror and in front of it a pointobject O is
Geometrical Opticsj
placed of which thepointrealimage 7isproduced. Similarly in fig;ure5.91 (b)shows a convex mirror forming pointsized virtual image /ofthe pointobject O placed in frontof it.In general for any case ofgeometrical opticsit is a general trend to represent the distanceofobjectfrompole of mirror bya letter 'w' and the distance ofimage fromthe pole of mirror bya letter 'v' and focal length of the mirror can be denoted by the letter'/'. Afi
Important pointsof measurements aboutthis sign convention are
•
All distances are measured from the pole of spherical
mirrors.
• The origin of the Cartesian coordinatesystemis considered at the centre of the optical device (pole of spherical mirror in this case).
/
•
—
c
0
»
\
F
r «—^
• Distances measured toward right ofthe optical device are takenpositiveand distances measured towardleft ofthe optical device are taken negative as in general we consider positive xaxis toward right of origin and negative xaxis toward left of origin.
(a)
• For the size of object above principal axis its height is considered positive and below the principal axis it is considered negative.
In figure5.91 (a) all distancesu, vand/are taken negativeand in figure5.91(b) distance u is taken negative and distances/and v are taken positive.
Figure S.9I
Note : In cases of spherical mirrors both of the above sign conventions gives same signs for the situations when object is
Important points of measurements aboutthis sign convention
placed on the left of mirror facing its reflecting surface and different signs when object is placed on the right of mirror facing its reflecting surface.
are
•
All distances are measured from the pole of spherical
.
5.9 Analysis ofImageformation by Spherical Mirrors
mirrors.
• Distances measured in the direction (or along) ofincident ray are taken positive and those measured opposite to the direction ofincidentray are taken negative.
• For the size of object above principal axis its height is considered positive and below the principal axis it is considered negative.
In image formation by spherical mirrors for a specific object, we mainly analyze four specific characteristics of image which includes ''Exact Location ofImage', 'Nature ofImage' (real or virtual), 'Orientation of Image' (erected or inverted) and 'Magnification ofImage' (enlarged or diminished with exact size ofimage). Lets discuss all these characteristics one by one. 5.9.1 Mirror Formula for Location ofImage
In figure5.91(a) all distances m, v and/are taken negative because these are measured from pole ofmirror in direction opposite to the incident ray and in figure5.91(b) distance u is taken negative and distances/and v are taken positive. Second Convention
Cartesian Coordinate System Sign
Mirror Formula is a mathematical relation between object distance, image distance and focal length ofa spherical mirror
from its pole. This formula is used in analj^is of image formation and used in finding the'Exact Location o/7wa^e'produced by a the mirror for a given object.
Convention
For the same situation shown in figure5.91(a) and 5.91(b) in this sign convention we consider a simple two dimensional coordinate system associated with the given spherical mirror for all measurements of distances.
Figure5.92 shows a concave mirror in front of which a point objectis placed at a distance u for which point image/is obtained at the location shown in figure. In the figure only one light ray from object is taken which incident at point and after reflection
IGeometrical Optics
215
it passes through I. The line joining point X and center of curvature of mirror C is the normal to the mirror at point X at which angle ofincidence and angle of reflection is same.
Note: It isadvisable to alwa>^ drawraydiagramwhilesolving a questionof reflection by a spherical mirror to get the proper understanding of the given situation and image formation by reflected light rays. Also students must note that above relations ofMirror formula is only valid for paraxial ra>^.Also students must note that above relations ofmirror formula is only valid for paraxial ra}^.
5.9.2 Analyzing Nature of linage Produced by a Spherical Mirror Figure 5.92
Forparaxial rays inthegiven figure wecanconsider angles 0], ©2 and 63 are verysmall so we can use h
h
h
a, = — ; a, = "T and a, = — ^
u
R
V
Now from ACXO we have ttj = aj + 0 and in ACXI we have a3= 02+ 0 and from theserelations wehave
By mirror formula when we analyze the location of image then with the location we get a clear idea whether reflected rays are converging or diverging. If image is produced in front of the mirror then reflectedrays willbe converging and image produced is real and ifimage is produced behind the mirror then reflected rays will be diverging and image produced is virtual. Figure5.93 shows the situation of real and virtual image formation by converging and diverging raj^.
2a2 = a[ + a3
Now substituting the Values of
a2 and
Spherical mirror (Concave or Convex)
in terms of h we
get
i R~ u
V
Real image
As R=f12we can write •" ^
1i
1
f
V
u
...(5.10) Spherical mirror (Concave or Convex)
Above equation(5.10) is called 'MirrorFormula' which is valid for both concave and convex mirrors and in this relation values
ofu, Vand/are substituted with proper signs according to any ofthe sign conventions stated in article5.8.8. In this book we will use Cartesian Coordinate S)^tem Sign Convention whereas in many books New Cartesian Sign Convention is used but it hardly makes any difference from student point ofview. For all the questions we are going to take up in this book, students can also try solving these questions by the first sign convention also for getting a clear picture ofanalysis. Equation(5.10) can be rearranged in three forms and used as per requirement for calculation ofw, v of/as given below. uv
/=
...(5.11)
SpJct
Diverging reflected rays
r''
Virtual image
Polished face
Reflecting face Figure 5.93
5.9.3 Magnification Formula for Size and Orientation ofImage Figure5.94 shows the image produced for an extended object (candle) by a concave mirror. In this case we are using ray1, ray2 andray4 as mentioned in article5.8.4 for image formation.
u + v
Now in this figure ifwe analyze in AAPBand AAPB'for paraxial
v =
ijf uf
...(5.12)
u
V/ V/
...(5.13)

Polished face
Reflecting face
rays we can get the relation in height of object hQ{AB) and height ofimagehfA BO perpendicular to principalaxis. hi
...(5.14)
1216
GeometriGa! Optics
real object or virtual image ofvirtual object) then image will be produced on other side of principal axis (w is negative) and if nature ofobject and its corresponding image produced by any optical device is opposite (virtual image of real object or real image of virtual object) then image will be produced on the same side ofprincipal axis (w is negative).
5.9.5 Longitudinal Magniffcation of Image Figure 5.94
Lateral magnification formula forspherical mirrorgivestheimage height above the principal axis ofmirror and in this section we will discuss on how to calculate the image width along the
Magnification ofimage is given as Height of Image m=
Height of Ojbect
...(5.15)
from equation(5.14) we have h; m
...(5.16)
=
Equation(5.16) is called''Lateral Magnification^ ofthe image which relates the objectand image height with their locations. Ifmagnification ismore than1then image formed will beenlarged and ifit is less than 1 then image formed will be diminished. For
m = 1 image height is sameas that of object height. Negative sign in equation(5.16) shows the orientation of image with respect to object. Ifin a case m is positive then it indicates that
principal axis of mirror. The relation in objectand image width along the principal axis of mirror is called 'Longitudinal Magnification^ as given below. Longitudinal magnification ofimage is given as m, =
Width of Image along Principal Axis Width of Object along Principal Axis
Figure5.95 shows image formation of an object located at a distance x from the concave mirror of focal length / which produces an image of this object at a distance y which is real inverted and enlarged because object was placed between F and C. Here we can see that object edge^ was close to C so
imageis formed on the samesideofprincipalaxiswhereobject corresponding image edge^'is also closer to C. Ifwe consider is located andifmisnegative thenit indicates thatimage formed objectis of verysmall width dx and imageproduced is having a is on the other side of principal axis where object is located. width dy then from mirror formula we have Thus for an erected objectif m is positiveimage produced by themirror will beerected andifmisnegativethenimage produced 1 =1 1 f
will be inverted.
X
y
Differentiating this expression we get
Lateral magnification formula canalsoberearranged in different ways usingequations(5.14), (5.15) and(5.16) for itsuses asper requirement ofthegiven parameters ina specific question. Below arethe given forms oflateralmagnification formula.
0=—\r dx—\dy X
y.
From this relation we can get the'Longitudinal Magnification' as
u
uf
f
...(5.17)
Note: As already discussed about the mirror formula that it is
m.
dx
...(5.18)
applicable onlywhen paraxial rays areforming theimage sothe magnification formula asexplained above will also bevalid only when paraxial rays are in consideration.
5.9.4 Relationin Nature and Orientation ofImage Forall the opticaldevices in geometrical opticswecan findthe orientation ofimage produced using themethod explained here. This method is onlyusing the basic geometry in finding the
B' .4'C Object
Figure 5.95
image orientationon principal axis. For small width object ifimage is produced by a spherical mirror
Always remember if nature of an object and its corresponding (concave or convex) then image width can be calculated by use image produced byanyoptical device aresame (real image of ofthe equation(5.18). But ifobject size is large then this relation
[Ge^efrical ^Opfics 217
cannot be used and in that case we need to calculate the image ofboth edges ofthe object along principal axis and take the difference as explained in figure5.96. Here we first need to
consider the distance ofedge5 ofthe object asx, and find the corresponding image ofthis edge at 5'located at adistance>'j
using mirror formula then we consider the distance ofedge^ of the object asx^ ~ +%and find the corresponding image^^ 'at location 3^2 using mirror formula. Now the width ofimaged 'fi' can be obtained as
(b)
Figure 5.97
In figure5.97(a) image edge I. can be obtained by lateral magnification as I. {where m= v/«) inboth sides ofthe square edges same I. will be there so final image produced is a
magnified squarereal image and similar to this in figure5.97(b) for convex mirror imagewill be adiminished square virtual image. Figure 5.96
The area ofimage produced is given as A, =If =nPll =m^A
In the same manner we can find the height ofthe image at the (^0" ^0 ofobject surface) edges and by separately using the lateral magnification at both theedges. CaseII: Object is placed in the plane ofprincipal axis 5.9.6 Superficial Magnification byaSpherical Mirror
Figure5.98(a) shows a small square object ABCD of edge ^0 (^0
which is placed at a point between F and C of a
As we have studied the lateral and longitudinal magnification, concave mirror ata distance u from the mirror. The image is we can easily find out the magnification in area for the image produced AB'CD'ata distance vfrom the mirror asshown and produced by a spherical mirror. There aretwo cases in which
in this case the image produced will be in shape ofarectangle
area ofimage is calculated which are given here one by one. In because for the edges BC and AD ofthe object which are both ofthese cases we are considering object size very small perpendicular to the principal axis, we use lateral magnification compared tothe distance of object fiom themirror. and for the edges AB and CD which are along the principal axis, CaseI: Object isplaced normal toprincipal axis
Figure5.97(a) shows asquare object
weuse longitudinal magnification.
which is placed at So in this case the image height and width are given as
a point between F and C of a concave mirror at a distance u
from the mirror. The image isproduced as^'B'C'D'at adistance v
h. = mlr.
from the mirror as shown and the image will also be ofsquare shape as both edges ofthe object are perpendicular to principal axis ofmirror so in both we use lateral magnification and size of both will be same. Similar situation for aconvex mirror isshown
w
where lateral magnification isgiven asm=~vlu
in figure5.97(b) 
^D'
C
T'lF "mi.
Cr' B'
A'
CA
Image
B
Object
B''
Image
(a) (a)
F
Geometrical Optics  i18.
„
+ve Virtual object
\ V> +ve Virtual image \
V
C'WWD'
\
/
/
>C^2/)
B' A' /
/ /
Real image , v ¥ +ve Virtual object u > ve
(/./)
/ /
/
*u
for Convex Mirror
(b)
Figure 5.98
u ) +ve v~¥ve
As Iq and w. are small we consider almost same height for image
Virtual object Real image
edges A'D' and B'Cand similar case for a convex mirror, is shown in figure5.98(b)
(b)
Figure 5.99
5.9.7 VariationCurvesofImageDistancevsObjectDistance
We have discussed themirror formula which relates theimage
#Illustrative Example 5.11
distance from pole ofmirror for a given object distance. The
Apoint object is placed atadistance 30 cm in front ofaconcave
mirror formula is given as
mirror offocal length 20cm. Find thenature and location of i_l
i
f
V
u
image obtained. Solution
v =
In this case for mirror formula we use / M=30cm
As focal length of a given mirror is constant and it can be
positive or negative depending upon the sign convention and type ofmirror. The above function can be plotted as shown in figure5.99(a) or(b) for a concave and convex mirror facing toward left along negative x direction which we generally
/=20cm
Usingmirror formula v =
consider.
• ~ve
• +ve
uf uf
+30X+20 30(20)
30x20
Virtual image Real object
v=
V
=60cm
Theimage formation is shown in raydiagram andwecan see thatimage produced isreal and located ata distance of60 cm
/ /
/ ✓ /
from the poleofmirrorin front ofit.
✓ ✓ ✓
' ✓
✓
 20cm
✓
fv^ ve
/ /
/ ✓
/
✓
U\3 ) 2/)
/ /
+ve
for Concave Mirror
\ \
u —> VA V—> —ve \
30cm
Real image) 60cm
Real object
(a)
Figure 5.100
[Geometrical Optics 219
# Illustrative Example 5.12
Solution
Athin rod oflength c//3 isplaced along theprincipal axis ofa For image produced by concave mirror, bycoordinate sign concave mirror offocal length = t/suchthatitsimage, which is convention for mirror formula, we use real andelongated, justtouches therod. Find thelength ofthe u = + 35cm,/=i25cm
image ?
Now by mirror formula we have Solution
uf
35x25
uf
10
= + 87.5 cm
v =
Given that image istouching theobject that means this point Thus imageproducedbyconcavemirror/is locatedat a distance 87.5cm from Pwhich isat87.5 —35 =52.5cm from the object O.
must be at the center of curvature of the mirror as shown in
figure5.101. The image of^ is formed at the same position (since is at Center of curvature). For image of 5 we will use
87.5 cm •
mirror formula.
35 cm
Centre of curvature
Separation between O & /is
Figure 5.101
= 87.535 = 52.5 cm
Applying mirrorformula forimageoiB, according tocartesian coordinate signconvention, weuseu = +Sdl3 and/= + d
Position of
from object is at O distance 52.5
= —^ =26.25 cm
i.l=l V u
•52.5 cm'
Figure 5.103
Distanceof
f
from poleof mirror is P = 35+26.25 = 6I.25 cm.
v'^ 5d
d
# Illustrative Example 5.14
5d
An observer whose least distance of distinct vision is 'd,
v =
views his own face in a convex mirror ofradius ofcurvature V.
Hence length ofimage obtained is
Prove that the magnification produced can not exceed
A'B'=—2d =^ 2
2
d +\ld^ +r^
# Illustrative Example 5.13 Solution
In figure shown find the distance from pole P of the concave
For clear vision, the distance between object andimage 01 mirror shown in figure5.102, at which when a plane mirror is mustbe morethan d. Ifin the figure shown distance OP bex placed, image produced byboth mirror for the object O will then by mirror formula, we have
coincide.
#
/=25 cm
M, 35 cm
Figure 5.102
I" Figure 5.104
^
Geometrical OpticsJ
......
Here bycoordinate signconvention, for maximum magnification
plane mirror M2 fi'om the object is30cm. We know that in a
we use 01 = d, so we use
plane mirror theimage isformed behind themirror at thesame distance as the objectin front of it. It is also giventhat there is
v = +{dx);u = x;f= + (forjc>0)
^
1
1
2
d~x
X
r
xd+x
2
{dx){x)
r
no parallax between the images formed by the two mirrors, thus the imageproduced by the convex mirror is formed at a distance of 10 cm behind the convex mirror which will coincide
with the imageproduced by plane mirror at a distance 30cm behind at position Q as shown in figure below.
 2jt^ + 2dx = 2rx —dr
Forming a quadratic in x we have 2 values
dr±yjr^ +d^ x
=
h«10cm
discarding the negative value we have
dri\lr^+d^ • x
=
Figure 5.105
Forreflectionat convexmirror using coordinate sign convention for mirror formula we have
M= 50cm, v= 10cm
Magnification produced by the mirror is
m
=
—
dx d +r~ylr^ +d^ dr +ylr^ +d^
Using mirror formula, we have
1 =1 +1 f
Rationalising the above equation, we get
m =
30cm
20cm
J_ _ J
(d +r){^lr^+d^f {dr +^r^ +d^ ){d +r +ylr^ +d^) 2rd
u
V
I_ _
/ " 50 10 "7 50 50
So radius of curvature ofthe mirror is
2d'^+2d{4r^+d'^) Thus the maximum magnification produced by the mirror can be given as
d +yjr'^ +d^
4
\
50x2
R = 2f= — c m the radius of curvature ofconvex mirror is 25 cm.
^ Illustrative Example 5.16
Using a certainconcave mirror, the magnification is found to # Illustrative Example 5.15
An object is placedin fiont of a convex mirror at a distance of 50 cm a planemirroris introduced in fiont ofthe convex mirror
be 4 times as great when the object was 25 cm fi:om the mirror as it was with the object at 40 cm from the mirror, the image, being real in each case. Find the focal length of the mirror.
covering the lower halfofit. Ifthe distance between the object
Solution
and the plane mirror is 30 cm and it is found that there is no parallax between the imagesformedby the twomirrors for the same object. What is the radius of curvature of the convex
We use magnification
mirror?
V m= 
u
j
Solution
Let O be the objectplaced in front of a convexmirror A/, at a distance of 50 cm as shown in figure5.105. The distance ofthe
/ =
7
w/
In first case when u = 25 cm, magnification is m
•
/ 25/
...(5.19)
jGeometrical Optics
22/]
In second case when m= 40 cm, magnification is
.=36
Now using mirror formula we have
J
...(5.20)
1_ J_
15~70
V,
As itisgiven that w, = 4ot2 we use / 25/
/ 40/
= 4
c , • we get Solving
v, = 210 cm
Thus final image / is formed in fi"ont of concave mirror at.
Solving we get/= 20cm.
distance of(210/l 1) cm and it is real.
# Illustrative Example 5.17
Magnification Wj for first reflection is
Aconcave and aconvex mirror each 30 cm in radius are placed opposite toeach other 60 cm apart on the same axis. An object 5cm inheight isplaced midway between them. Find the position and size of the image formed by two successive reflections,
12. 30
Magnification
for second reflection is
consider first reflection at convex and then at the concave mirror.
Solution
V2
(210/11)
~ «2 ~
70
3
"IT
Final magnification = w, x w, = —x— = — ' 2 3 11 11
The ray diagram of image formation is shown in the below figure5.106.
Thus size ofthe final image=5x^ ^ cm. 5.9.8 Effect ofMoving Object and Spherical Mirror on Image When object ormirror isinmotion thedistance between object and mirror changes which affects the position and size ofimage. To find the image velocity and for analysis ofimage's motion we can diflferentiate the mirror formula and find the rate at which
r = 30cm
r  30cm
Figure 5.106
distances between object orimage and mirror ischanging. Ifwe consider a; and y as object and image distance fiom pole of mirroroffocal length/then bymirrorformula wehave
For first reflection at convex mirror, using coordinate sign convention we have for mirror formula
Mj=30cm, / = +15cm
1
f.
y
X
Differentiating the above relation with respect totime, we get
Now using mirror formula we have 1
11
0=J .
^ dt
• 1
1
dt
•+ —
15 or
30
Vj =+ 10cm
dx .
Where — is therelative velocity ofobject with respect tothe
Thus image produced will be virtual. This is formed at/ behind the convex mirror at adistance of10 cm. The image /, acts as a mirror and — isthevelocity ofimage with respect tomirror. real object forconcave mirrorbecause its diverging light rays fall on the concave mirror. X
Now for second reflection atconcave mirror, using coordinate sign convention for mirror formula, we have 10 = +70cm, / = + 15 cm
v, = 
y 2
...(5.21)
Geometrical Optics 
Where m is the lateral magnification produced bythe mirror.
Ifpolice jeep is at a distance behind the thiefscar then we
Theexpression ofimage speed as given in equation(5.21) is valid onlyfor the velocity component ofthe image and object along the principal axis. If the object and mirror is in motion along the direction normal to principal axis we can directly
can use m =  c/ so we have
differentiate theheightofobject andimage above principal axis
^
which are related as
Thus distance of image from mirror is
10
1
10M)
10
t/=90m
h: = mh v = mu =
Differentiating this with respectto time we get dh;
= 9m
Now rate at which magnification is changing is given as
dh„
— = m —T"
dt
dv
dt
j
dm v,^=wv ON
Here we can use
dh
=v.^and
u
du V—
dt
dt
dt
d\
=VQ^which are the velocity
components ofimage andobject respectively in direction normal to the principalaxis. ^
dm
(90)(lxl0^)9(l)
~dt'
90^
dm
81
~dt
10x8100
ni
= +l X 10"^ s"'
§ Illustrative Example 5.18
it Illustrative Example 5.19
A thief is driving away on a road in a car with velocity of 20 m/s.Apolicejeep is chasing him,whichis sightedbythief A Convex mirror ofradius ofcurvature 7? is fixed on a stand at in his rear view mirror, which is a convexmirror of focal length rest with total mass m which is facing a block ofequal mass m 1Om. He observesthat the image of the jeep is moving towards as shownin figiire5.107 and iskepton a fiictionless horizontal him with a velocity of 1cm/s. Ifthe magnification ofthemirror surfece. Theseparation between theblock and mirroris IR and block is moving at a speed vtoward the mirror. Consider elastic for the jeep at that time is 1/10. Find: collision between block and stand, find the speed of image
(a) The actual speed of the jeep.
(b) The rate at which magnification is changing.
after time 3i?/v.
Assumethat policejeep is on axis of the mirror. Solution
(a) The velocity of imagewithrespect to mirror is related to velocityof objectwith respect to mirror is given as
IR
Figure 5.107 Solution
=>
 1 X 10"^= After elastic collision block will come to rest and mirror with
stand starts moving at the same speed v and the collision occur at time IRIv and after a fiirther time RJvwhen total time elapsed
Velocity of object with respect to ground is given as Vn,n=Vn,^¥V.ml G
is 3i?/vtheseparation between blockandmirrorwill be R and at this instant the position of image will be given by mirror formula with
Vo/a = ^+20=(+21 m/s)/ (b) The magnification produced by the mirror is m =
/ fu
1 10
u=RJ=+RJ2 Now using mirror formula we get
uf _ ~Rx{RI2) _ R v =
uf ~ ^RRI2 ~ 3
fGjeometrical Optics
2231
At this instant magnification is
^
V _ 7g/3 ~ u
Same effect can be there on image ifa spherical mirror iscut and displaced. In figure5.109(a) aconcave mirror isforming image/ ofan object O. If it iscut atcenter andthetwo parts and aredisplaced indirection normal toprincipal axisbya distance Xas shown in figure5.109(b) theneach part ofthemirror will behave like a separate mirror and produces its own separate image ofthe given object. Due tothis cutting and displacement ofmirror parts, object distance from each ofprincipal axis of these parts has become x/2 and by these two mirror parts two
1
^~3
Speed of imagewithrespect to mirroris given as
/i\2
il v=i 3
9
V
images are produced at a distance y above and below the two principal axisin thesame plane ofprevious image as shown in figure5.109(b) wherey isgiven as
10
Speed of image with respect to ground is v + —= — v
5.9.9 Effect ofshifting Principal Axisof a Mirror
y=
.X
Figure5.84 shows a point object and its corresponding point image produced by a concave mirror. If in this situation the
mirror is displaced upward bya small distance Ax in upward direction then with mirror its principal axis will also shift as
cut here
shownin figure5.108 and in final state object willbe located at a distance Ax below the principal axis of mirror. Due to this image will be displaced up and finally obtained above the
principal axis at a distance Ax + Ay from its initial position. Here (a)
Ay can be directly given as Ay = w Ax
X
(b) Figure 5.109
(a)
Note: If in figure5.109(a), a small section at its center of width
Xiscutand removed as shown infigure5.110 then image will remain at the same position where it was because now the two
parts ofmirror will behave asa single mirror with same principal axis.
(b)
Figure 5.108
Figure 5.110
Geometrical Optics t
\22A
5.9.10 Image formation of distant Objects by Spherical Mirrors
incident rays offigure5.112(a) andmeet atpointO,andproduce theimage I2 at this location as shown.
For distantobject, weknowthat a sphericalmirror produces its image in focal plane. The size of image can be calculated by analyzing its angular width with respectto the pole of mirror. Figure5.111(a) shows the formation of image of Sun by a concave mirror. S is the solar disc and here 6 is the angle
subtended by solar disc at any point on earth surface which will be same at the pole of mirror as shown. Here we consider twolight raysfromthe edgesofthe solardiscwhichincidentat pole of mirror at angle 0 and these get reflected at the same angle and producethe real image of Sun on focalplane. Here we can see that the image produced is inverted and its diameter can be given as =/0 as 0 is a very small angle and image is obtained at a distance/from the pole. Figure5.111(b) shows the formation ofimage ofthe Sun by a convex mirror. The only difference here is that virtual erected image is produced at focal plane as shown. A
(b) Figure 5.112
•
S/
Here we can state ifan object is placed at the location ofimage produced by a spherical mirror then for this position of object, image is produced at the location of initial object.This concept is called 'Reversibility ofLight' for image formation by mirrors. Real image X 1^
# Illustrative Example 5.20 •/—H
A concave mirror of focal length 20 cm is cut into two parts from the middle and these two parts are moved perpendicularly bya distance 1 cm fromthe previousprincipal axis/fS. Find the distance between the images formed by the two parts ?
(a)
Virtual
10cm
image
Figure 5.113
(b)
Figure 5.111 Solution
5.9.11 Concept ofReversibility of Light
Figure5.112(a) shows an object in front ofa concave mirror of which image is produced at point In this case we have
discussed that allparaxial lightra>^ from O, gets reflected from themirrorand afterreflected raysintersect at point/, andproduce image oftheobject. Ifweplaceanother object O2 attheposition /, as shown in figure5.112(b) then according to laws of reflection all light rays from this object will incident on the mirror following the same path offigure5.112(a) in opposite direction and gets reflected from mirror and follow the path of
Consider the figure5.114 in which we take
is the principal
axisforthemirrorM, &P2 is theprinciple axisfor themirror M2 and/,, I2are the imagesproduced bymirrorsM, and A/j. Using coordinate convention for mirror formula here we use w= 10cm and/ = 20cm Now by using mirror formula, we have
=>
1
J_ _ J_
10
V ~ 20 V = + 20 cm
'Geometrical Optics 225
Thus image will be produced by both mirrors at a distance Web Reference at www.phvsicsgalaxv.com
20cm behind the mirrors as vis'+ ve'. Now the magnification produced ism= 20/(l0) =+ 2thus image will be produced on the same side ofprincipal axis ataheight 2(1 cm) =2cm from
Age Group  High School Physics ] Age 1719 Years
the principal axis.
Topic  Geometrical Optics IReflection ofLight
Section  OPTICS
Module Number  24 to 40
Figure5.114 below shows the two images /, and /j produced by the two mirrors. Practice Exercise 5.2 10cm
P,
/,_
Q
A
20cm
2cm
P
(i) Find the distance from a convex mirror, offocal length 60cm where an object ofheight 12 cm should be placed sothat
its image is produced at 35 cm from mirror. Also find the height ofimage. [ 84 cm, 5 cm]
Figure 5.114
The heights of the two images are given as
hj 2 cm below the principal axis P,
and hj^ =2cm above the principal axis Thus the separation between the images is given as
(ii) A 2cm high object is placed on theprincipal axis of a concave mirror ata distance I2cm from the pole ofmirror. Find
the location ofthe image and focal length ofmirror ifthe image height is 5cm and it is inverted. [30 cm, 8.6 cm]
7,72=2cm (Hi)
# Illustrative Example 5.21
An object is located at a distance 30cm onprincipal axisfrom the pole ofaconcave mirror offocal length 20cm. Suddenly the mirror is displacedby a distance 1.5cm in the direction normal
to its principal axis. Calculate the displacement of image
A pointobject on theprincipalaxisofa concave mirror
offocal length 20cm, ismoving ata speed of5cm/s atan angle 45° to the principal axis asshown in figure5.115. Initially object islocated ata distance of25cm from thepole ofmirror. Find the velocity components ofimage along and normal to principal axis at this instant.
produced by the mirror due to this.
/= 20cm 5cm/s
Solution
^5° By using coordinate convention if we consider the object to be located to the left of the mirror, we have
25cm
w= 30cm and /=20cm Figure 5.115
By mirror formula we use v/ i/ =
vf
Magnification is w =
(iv)
30X20 10
=60 cm
60
A man uses a concave mirror for shaving and sees his
2 timesenlarged imagein mirror when his faceisat a distance 40cmfrom mirror. Findfocal length ofmirror. [80 cm]
30
If mirror is displaced up by a distance 1.5cm then object
(v)
distance from principal axis of the mirror becomes 1.5cmand
concave mirror ata distance of12 cmfrom thepole. Ifthe image is inverted, realand5cm high, find thefocal length ofmirror. If
the new position of imagewillbe at a height2 (1.5)= 3cm
from the principal axis which was earlier produced on principal axis.
Thus displacement of image is 3cm.
A 2cm high object is placed on the principal axis of a
the object starts movingat a speed of 1.2cm/s toward the mirror find the speed ofimage and its direction of motion. [7.5 cm/s]
Geometrical Optics
'226
(vi) Figure5.116 shows two spherical mirrors Mj and on same optical axisat a separation of50 cm. Apointobject O is placed midway between mirrors onoptical axis. Find location & nature of its image after two successive reflections first at A/, then at Mj. •/f^ = 20 cm
Denser medium
.
= 30 cm
Rarer
medium
Angle of deviation 5 = 02  0 (b) Figure 5.117 f*
50cm
,
H
5.10.1 Absolute Refractive Index of a Medium Figure 5.116
When a light ray enters a mediumthen how its speed changes is measuredby ^Refractive Index" of the medium. It is defined as the ratio ofspeedof light in the medium to the speedoflight in free space andthis is alsocalled absolute refractive indexof
[75 cm]
the medium.
5.10 Refraction of Light
Absolute Refractive Index ofa Medium
Whenever a lightrayis incident on theboundary oftwodifferent media, a part of it is transmitted into the second medium and whenever a light enters in a transparent medium, it suffers a sudden change in velocityof light which we call refi^action of light. In the process ofrefraction the direction of propagation oflight ray changesifit incidentson the surfaceat someangle of incidence to the normal to boundary as shown in
Speed of Light in free Space Speed of Light in the Medium ...(5.22)
Wherec is the speedoflight in free space and v is the speedof light inthegiven medium. Thus themedium inwhich light travels
figure5.117(a). This case corresponds to the situationwhen a faster has lower refractive index and is called 'Rarer Medium'" light ray travellingin a rarer mediumincidenton the boundary compared to another mediumin whichlight travelsslower and of a denser medium after entering into the denser medium light has higher refractive indexwhichis called'Denser Medium'. ray bends toward normal, which you might have covered in yourprevious classes also. Figure5.117(b) showsthe situation 5.10.2 Relative Refractive Index of a Medium when a light ray travellingin a densermedium incidenton the Manytimerefractive index ofa medium is specified asrelative boundary of a rarer medium and after entering into the rarer refractive index with respect to another medium. This can be mediumlight ray bends awayfromthe normal as shown. defined by the relation similar to equation(5.22). Relative refractiveindex ofa medium1 with respect to another medium2 is given as Rarer
Relative R. 1ofmedium1 with respect to medium2
medium
Speed of Light in Medium  2
Speed of Light in Medium1 Denser
medium
...(5.23)
Here 2M.1 isthe way how we denote refractive index ofmedium1 Angle of deviation 5=010,
with respect to medium2. So if we define refractive index of medium2with respectto medium1,we write 1
(a)
11^2'
21^1
Geometrical Optics 227
Here for the two media if their absoluterefractive indices are
taken as and
we can write )x, = — and ji,= — V2
to medium2 the product ofrefractive index ofthe medium'and
thesine ofthe angle made by thelight raywith normal in that medium remain constant. In this case we have
p,sin0, = P2 sin02 =Constant
5.10.3 Laws of Refraction
When a lightraypropagates through aninterface oftwo different media then its behaviour is governed by certain laws called lawsofreffaction which are also known as 'Snell'sLaws'. There are two laws ofrefraction which are described here.
...(5.24)
Even for several media separated byparallel interfaces asshown in figure5.119, the same relation holds valid.
(1) sinOp = pjsin0[ = p^sinGj^ p3sin03 = p^sin0^^ ... = constant
First LawofRefraction : In therefraction ofa light rayat a medium interface the incident ray, refracted rayandnormal to theinterface lieinsameplane called 'Plane ofIncidence' onthe
medium interface. Figure5.118(a) shows the incident ray, refracted rayand normal and inthis figure the plane ofpaper is considered as plane of incidence. Figure5.118(b) shows a perspective view oftherefraction ofthesame lightrayinwhich plane of incidence is shown by the grey shaded plane which
Medium!
Ml
Medium2
M2 x!
03^ Medium3
makesthe understanding better. :.Medium4
^4
Or..
'co
Normal
%
Figure5.119
Medium1 (ji,)
Asshown in above figure wecanseethatforparallel interfaces
Medmm2 (ji,)
when light comes out inairagain it comes out atthesame angle 00 andemerges outparallel to theincident ray. 5.10.4 Vector form ofSnell's Law of Refraction
The second law ofrefraction relates the refractive index and the
(a)
angle light raymakes with thenormal inthatmedium. The angle can be expressed by using unit vectors in the direction of
incident ray, refracted rayand normal to" the boundary. If in figure5.120 we consider unit vectors /, r and n along the direction of incident ray, refracted ray and normal as shown Plane of Incidence
then the equation(5.25) can be written as
p, (/ X«) = p^(rx it)
(b)
FigureS.118
Second Law of Refraction : For the situation shown in
figure5.118(a) fortherefiaction oflightraygoingfrom medium1
Figure 5.120
...(5.25)
Geometrical ^OptlcsJ
228 .
Above equation(5.25) isused for analyzing law ofrefraction in vector analysis whenever in different questions light raysare specified in vector form orvectors aregiven along thedirection of incident and reflected rays.
5.10.5 Image Formationdue to Refraction at a Plane Surface
When all lightrays from an object falling onthe planeinterface of two different media as shown in figure5.121 these all rays are refracted in such a way that after refracting these rays will
appear tobecoming from a point / which istheimage ofobject as seen bv the observer shown in figure.
Dividingequation(5.28) byequation(5.27) we get ...(5.29)
Equation(5.29) iscalled asrefraction formula for image formation by refraction oflight at plane surfaces. Whenever an object is placed at a distance u from a plane interface oftwo transparent media then theimage ofobject asseen from theother media will appear at a distance v as shown in figure5.121 which can be obtained by using this formula given in equation(5.29) and thisimage distance isalsocalled Apparent Depth' ofthe object. 5.10.6 An Object placed in a DenserMediumisseen fromAir
Medium2
Medium1 1^1
Observer
Objccl
Figure5.123 shows an object placed inwater(denser medium) having refractive index p anditis seen byanobserver from air. Ifobject isplaced ata depth hbelow theairwater interface then the image of the object is seen.by the observer at a depth h' (apparent depth) givenby refractionformula as l^l
^2
Imase
Observer
Figure 5.121
If we consider a specific paraxial ray from object incident as shown in figure5.122 which incident on the interface at an angle 0 and gets refracted at angle (jithen by Snail's law we can
water (u)
write
...(5.26)
p, sin 0 = Pj sincj) Medium1
Medium2
Object Figure 5.123
Here w= /?, v=
p, = p and Pj = 1sowehave h__h
It" 1 Figure 5.122
=>
h'=
...(5.30)
P
Forparaxial rayson the interfacewe can considerangles0 and (f) are very small so we can use sin 0 ~ 0 and sin ({) ~ (j) so from
If the distance on the interface AP is taken as r then for small
Equation(5.30) is the relationused for 'Apparent Depth ofan object placedin a densermedium seenfromair. In figure5.123 we can see that object appear to be closer to the interface compared to the actual depth of the object which generally happens when we see an object placed inside water level at some depth. The shift of object due to refraction can also be
angles 0 and (J) we have
calculated and given as
above equation(5.26) we have •
Pie=P2()
r = iiQ = v(j)
...(5.27)
... (5.28)
Shift
h AS = h   = h P
11
...(5.31)
jGeonietricai Optics' 229
Above equation(5.30) and equation(5.31) are valid only for # Illustrative Example 5.22 paraxial rays. In case ofplane surface we consider it only for
observation ofobject along the normal or with the light rays at aconvergingbeam oflightrays incident on aglassair interface near normal incidence. ^^own in figure5.125. Find where these rays will meet after S.10.7 AnObject placedinAir and seenfroma DenserMedium
refraction.
Figiire5.124 shows an object placed in air having refractive and itisseen by an observer from water (denser medium) with
glass (P = 3/2)
refractive index p. Ifobject isplaced ata depth h above theairwater interfece then the image of the object is seen by the observer at a depth h' (apparent depth) given by refraction
20 cm
Virtual object
30cm
formula as u
V
Fi
F2
Figure 5.125 Solution
* Image /f'v
/' ' I I
For plane surface we can use refraction formula as
«\ I \
t\ \ Object T
ilj V,
h'
' / /
' /'/ ij'i ir't
'/ '
"
\ N \ \
I 5
1 iair(l) water (ji)
F2
2
i/ = + 20
/ /
B.
h.
u
V
1
3/2
20 "
Observer
V
v = + 30cm
# Illustrative Example 5.23
Figure 5.124
Aconcave mirror isplaced inside water with itsshining surface
HereM = /i, v= h\ pj= 1andp,  ft sowe have
upwards andprincipal axisvertical as shown in figure5.126. Rays are incident parallelto the principal axis of the concave mirror. Find theposition ofthefinal image.
A! 1
=>
ft
h'=\ih
...(5.32)
Equation(5.32) isthe relation used for'Apparent Depth' ofan object placed,in airasseen from adenser medium. Infigure5.95 we cansee thatobject appear tobedisplaced farther away from the interface compared to the actual depth of the object this generally happens when an underwater diver sees an object
Air Water 4/3 •'Ocm
placed outside in air level at someheightabove thewaterlevel,
R = 40 cm
the object appear tobe further away from water level compared
Figure 5.126
to its actual height. The shift of object due to refraction can also be calculated and given as Solution
Shift
A5=pA/i = /i(pl)
...(5.33)
The incident rays will pass undeviated through the water In thiscase also above equation(5.32) andequation(5.33) are surface and strike the mirror parallel to its principal axis. also valid only for paraxial rays as these are obtained bythe Therefore for the mirror, object isatoo its image/J (in figure5.127) refraction formula which we derived using lightrays with small angle of incidence or near normal incidence.
will be formed at focus which is 20 cm from the mirror. Now for
the interface betweenwater and air, d= 10cm
Geometrical Optics
!230
# Illustrative Example 5.25
t
Figure5.129 shows a concave mirror offocal length Fwithits principal axis vertical. Inmirror atransparent liquid ofrefractive index p isfilled upto height d. Find where on axis ofmirror apin
Water 10 cm 4/3
should beplaced sothat itsimage will be formed on itself.
30 cm
^ = 40 cm
Figure 5.127 d
d'=r^
10 = 7.5 cm
4/3  1
# Illustrative Example 5.24
Figure 5.129
Abird in air is diving vertically over atank with speed 6cm/s. Solution The base ofthe tank is silvered. The fish in the tank is rising
For image of prism to be produced on itself we use
upward alongthe sameline with speed 4 cm/s. (Take: ^^=4/3). Find:
h\\.+ d=R
(a) The speedof the image of the fish as seendirectlybythe
R = 2F
bird.
2Fd
(b) The speed of the image of the bird relative to the fish
h It
looking upwards. # Illustrative Example 5.26
Solution
(a) Velocity offish in air = 4x—=3t
The^Tplane is the boundary between two transparent media. Medium1 with z > 0hasa refiractive index V2 andmedium2
Velocity offish w.r.t. bird = 3+6=91
with z < 0 has a refractive index
A ray oflight in medium1
given by the vector ^ =6VJ + 8>/3 j 10^isincident on the planeof separation. Findthe unit vector in the direction ofthe refracted ray in medium2. Solution
See figure5.130
Figure 5.128
I
Figure 5.130
(b) Velocity ofbird in water =6x^ =8i Given that
w.r.t. fish = 8 + 4=121
AB = 6^/3/ + sVsj'lO^
IGeometrical Optics From figure
231
consider object is placed at a distance h from the front face of
AB = AM + MB
Thuswehave
= eVJf + SN/Ijand ^ =10/t
theslab, first image /j due tofirst refraction will beproduced at a distance \ih from the surface as shown which is given by
The angle between the two vectors can be obtained byusing
equation(5.32). If glass slab thickness is t then for the second
the following formula
refraction image will act likean object located at a distance (/ + p/j) andfinal image/2 isproduced at a distance (t + \xh)/^ at a distance from this face of glass slab which is given by equation(5.33). Sofor the observer looking at the object from the other sideof glassslab the shift of object observed will be
ABMB cos 1 =
_AB\\MB\
Hence
^= I
11A  cos;
given as
10 cos / =
V36x3 + 64x3 + 100
t + \xh
Shift
2
F
cos i = —andsin i = V3 / 2
11
...(5.34)
By using Snell's law, we have
...(5.35)
Equation(5.35)is the expressionused for calculation of shift of objectdue to refraction by a parallel sided glass slab.
a, sin 1= ^2 sin r
=5«
V2 sin/ = >/3 sinr
=>
^/2x(^^/2) = V3 sinr
glass (u)
• 1 and^ cosr=—7= 1 => r = 45® smr=7=
=»
^/2
consider vector BN.
BN =fiA'^cosr( k) + BNsinre where e is unit vector along CD
.
bVJffsVJ?
10V3
6t
8,
"io'"'io^
unit vector along reflected rays is —
5D
1
.
1
—i H 10
1
I0V2 _
1
" 10^^ V2
j 10
(a)
1
(10/fc)+7^7j= (6/+8y)
IOV2
[6/ +8y10;t] „
.
.
^^[3/ +4J~5k]
Final Image
(b)
S.10.8 Shift of image due to Refraction ofLight by a Glass Slab
Figure 5.131
When a light ray from an object O is incident on a glass slab at some angle / it gets refracted inside the slab at angle r and from its other face it emerges out in air at the same angle i as shown in figure5.131(a). In the whole process light ray suffers two refractions at the two parallel surfaces of the glass slab. If we
Figure5.131(b) shows the ray diagram of formation ofimage which is seen by an observer from the other side ofglass slab. The expression ofshift ofobject as given in equation(5.35) is also valid only for near normal viewing as the result is obtained for paraxial ra>^ only.
Geometrical Optics t
232
5.10.9 Shift due to Refraction of Light by a Hollow thin walled Glass Box placed inside a Denser Medium
Figure5.132 shows a hollowboxmadeup ofthin wallsofglass is placed in water and a light ray from an objectO falls on the first interface ofthe box in which the light enters into air from water as very thin walls of glass can be neglected because these thin walls of glass will not produce any shift or lateral displacement in path oflight ray.
5.10.10 Lateral Displacementof Light Ray by a Glass Slab
Figure5.133 shows therefraction ofa lightraythrough aparallel sided glass slab which incidents on the slab at an angle of incidence z and enters into the slab at angle of refraction r
finally emerges out in direction parallel to the initial ray. If refractive index of the glass used is p then at the point A in figure we can apply Snell's law as ...(5.38)
sm z = p sm r
Water (h)
A
0
/,
11
^
r air(l)
Figure 5.133 h
Hi /+
In the above figure in right angled triangle A/4CD we can use
M
....
As we can see in the above figure, a light ray from object O located at a distance h from the box gets refracted from first face
ofbox andproduces an image/, at a distance
hoUow ho.x
/+—
Lateral Displacement oflight ray
A^= CD ACsin(ir)
which willact
as an object for second refraction at second face of box and finally the image is produced at a distance m{t+ /i/p) which is seen by the observer looking at object located on the other side as shown. We can see that the final image of object appears to be shifted away from the box and this shift can be given as h
CD
smOr)=^
Figure 5.132
{h+t)
in bxAEC we havecos r =
t
AC t
...(5.40)
AC = cosr
From equation(5.39) and equation(5.40) we get lateral displacement oflight ray is given as r.sin(zz')
A5.hollow
Note : If the light ray from an object placed in a medium with
refractive index p, passes through a slab ofthickness t made up ofanother medium ofrefractive indexpj andfrom othersideof slab in the same medium of object an observer is situated and sees toward object then its final image appears to be located with a '5/z/y?' which is given by the general formula used for
cosr
Ifincidence angle z is very small (for near normal incidence), then we can use
sin(z'r) ~ ir cos r = 1
A = /(zr) For very small i, from equation(5.38) we can use
shift due to a parallel sided slab
z= pr
1_£l
...(5.37)
In equation(5.37) ifA5 is positive then shift is taken toward the slab and ifit comes negative then it is taken away from the slab. With a careful analysis of this formula you can get equation(5.35) and equation(5.36) by properly substituting the values
of p, and P2 in it.
...(5.41)
A =
...(5.36)
box
...(5.39)
(as for small 0, sin 0 = 0)
z
Thus in equation(5.41) for small angles we can write sin (z r) = (zr) and cos r = 1, we get A = ti
ii
Geometrical Optics 233
5.10.11 LateralDisplacementofaLightRay due to Refraction 5.10.12 Concept ofReflection bya Thick Mirror by Multiple Glass Slabs
Aplane mirror isathin glass slab polished atone ofitssurfaces
When several parallel sided glass slabs are placed adjoining to due to which the other side becomes reflecting. Ifthe glass each other then due to refraction while passing through these used in making mirror is thick then it also shifts the image before
slabs thelight rayisdisplaced laterally which can be calculated
reflection from therear polished surface aswell asafter reflection
bysummingup the individual displacement ofthe light which and before coming out in air. Light passes through the glass of occurs when the ray passes through the slabs independently mirror twice as shown in figure5.135{a). Figure5.135(b) shows as shown in figure5.134. athick plane mirror ofglass thickness /placed in front ofapoint object O. A light rayfrom Ofirst gets refracted from the front
glass face
then gets reflected from mirror A/(polished face)
and after reflection again gets refracted from the front face S^ before coming out in air and finally produces the image L. Here /] and/j are the intermediate images produced by the light ray.
(a)
Figure 5.134
Asshown infigure5.134 when a light rayisincident from airon
the surface ofthe first slab and ifonly this slab ispresent inair,
(h + t/ix)
the light will come out in air along the dotted lineL, as shown in
figure and the displacement ofthe light would have been Aj and ifadjoining next slab isplaced at a negligible separation with first slab then it enters into it and comes out along the dotted lineLj and the lateral displacement ofthe light ray would have been A, +A^. In the same manner ifthird ormore slabs are (b) Figure 5.135 placed all the lateral displacements produced by each slab independently for the angle ofincidence i are added together. If r,, r^, are the angles which light rayismaking with thenormal Final position ofimage produced can be directly calculated by in each ofthese slabs then the total lateral displacement oflight considering the shift inrear surfece as observed by the observer. Due tothickness ofglass the polished face appear to be at an ray is given by
apparent depth r/jiasshown in figure due towhich thedistance between object and this newposition M' of the mirror will be
Ay.— A + A2 + Aj +...
(h +t/\x) and now the image will be produced exactly atthe same distance behind the mirror. So the separation between object
_ f.sin(/rj) r.sin(fr2)_ r.sin(fr3) =>
^^+ ^ ^+^
cosr,
C0S/'2
(5.42)
COSr3
Fornearnormal incidence oflight this equation(5.42) will reduce to the equation given below
Aj, /,/• 1— +v 1—+V
1—
.
F3.
+
.(5.43)
and image can be given as
Lor2\h^
...(5.44)
Students are advised to verify the separation given in equation(5.44) using calculations in stepped manner byfinding the locations ofimages /j and then step by step.
J2^4
Geometrical Optics
.
# Illustrative Example 5.27
Considerthe situation shownin figiire5.136. Aplanemirror is
fixed at a height h above the bottom of a beaker containing water ofrefractive index m uptoa height hj.Find theposition
Beaker
ofthe image ofthe bottom formed bythe mirror. Glass
Coin
Figure 5.137
^
h
o
+ t2 f,1
I
f
3^ +10
= 10 1—
l
^2^
4j
2)
[
1
3j
Hencethe coin appearsto raised by 5.8 cm. Now the apparent depth of the coin is 20  5.8 = 14.2cm. Figure 5.136 Solution
(b) As is evident from figure CD and EF are the refracting surfaces. At surface CD, the ray passes from glass to water and at surface EF, the ray passes from water to air.
Due to water in beaker the bottom of beaker appears to be
shifted upward and its apparent depth is given as h^lp. Thus the distanceof refracted image of bottomfromthe plane
mirror is/? /ij + hj\i. Plane mirror produces the image atthe same distance behind it so final image of bottom after reflection
from the plane mirror is produced at a distance hh^+ h^l\x
3/2
Case{i)
4/3
Critical angle at this surface is given by
C, = sin' (8/9) orC, =62M5' Case(ii) Similarly,
above the mirror.
# Illustrative Example 5.28
C2 = sin'(3/4)orC2 =48°36'.
A rectangular glassblock of thickness 10 cm and refractive As wesee that criticalangleis smallerin case(ii),total internal index1.5placed over a smallcoin. Abeakerisfilled withwater reflection occurs at the upper surface DF earlier as the eye is of refractive index 4/3 to a height of 10 cm and is placed over the glass block.
moved away from the normal.
(a) Findtheapparent position ofthe object when it isviewed
# Illustrative Example 5.29
at near normal incidence.
(b) iftheeye isslowly moved away from thenormal ata certain position, the object is found to disappear, duetototal internal
Aglass slab ofthickness 3 cmandrefractive index 1.5 isplaced
reflection. At what surface does this happen and why?
a point object be placed if it is to imageon to itself?
Solution
Solution
Let ABCD be a glass block, (p = 1.5) placed over a coin as shown in figure5.137. Let a beaker containing water upto a
Theglassslabandthe concave mirrorare shown infigure5.138.
height 10 cm placed over glass slab.
(a) As the ray from coin passes from glass slab to water it movesawayfrom the normal as it enters a rarer medium from a denser medium. Similarly at surface EF, the ray again moves awayfrom the normal. Whenviewed fromN, the coin appears
to beatly Apparent shift in multiple slabs is given by
in front of a concave mirror of focal length 20 cm. Where should
Let the distance of the object from the mirror be x. The slab causes a shift in position of object which is given as s t
ii
= 1 cm
The direction of shift is towards the concave mirror so the
apparentdistance ofthe object from the mirror is (a: 1) cm.
Geometrical Optics
235;
To produce the image onitselfthereflected rays must retrace Thus equivalent focal length ofthe combination mirror is given the path of incident rays so the object should appear at the as centre ofcurvature ofthe mirror.
/•
= 2
Slab
60 ~ 60
feq 30
Icm
0
1P
Je O
P
As image formed by the mirror liquid system coincides with
the source, the location ofobject isat 2f^ •R
60
^
.
w= 2/ =
Figure 5.138 60
Thus we use
x\ =2/= 40cm
According to the given condition, — + 30 is the distance of [I
theimage formed bythemirror itself, thus using mirror formula X = 41 cm from the mirror
we have
111
# Illustrative Example 5.30
V
A concave mirrorhasthe form of a hemisphere witha radiusof R = 60 cm. A thin layer of an unknown transparent liquid is pouredjnto the mirror. Themirrorliquid system forms onereal image and anotherrealimageis formed bymirroralone ofthe source in a certain position. Image produced bycombination coincides with thesource andthatproduced bymirror alone is
located at a distance of / = 30cm from the source away from mirror. Find therefractive index p ofthe liquid.
1 =>
u
f
Ixu
1
60
30
—I
V
''
60 v =
60
p2
Image distance from the mirror is
60
2p
, thus form the already
obtained condition we use 60
60
— + 30 = 
Solution
=>
\i 2p i^+ 2p4=0
For concave mirror with unknown liquid, equivalent focal
^l±V5
length of the combinedmirror is given as
J
2^1
•feq
fi
fhi
Where
Pi
p= I + v/s (as pcannot be negative) # Illustrative Example 5.31
Inprevious illustrations ifimage formed bythemirror coincides with the source and that produced by the combination is produced at a distance 30cm from thesource away from mirror then findthe refractive indexof the liquid.
60
A
Solution
Mirror produces its image on source when the source is located
at the centerofcurvature thussourceposition mustbe at 60cm from the pole ofmirror.
Now we use mirror formula for calculation of image distance Figure 5.139
for mirror liquid combination
1 +1 = ±
and focal length ofmirror is
^
"
fe
60
/«^Y=30cm
1 J_ V
60
iL 30
Geometrical Optics
236
60 v =
5
60
J(\0'+h^) _J_
12^1
1.5
According to the given condition we use 100+ /i^
 ^ +30 =60
25 +h^ ~ 9
2^l
:=>
16
h  8.45 cm
i=1.5
# Illustrative Example 5.33
# Illustrative Example 5.32
A person looking through a telescope Tjust sees the points
Asmallobject isplaced 20 cmin front ofa block ofglass 10cm
on the rim at the bottom of a cylindrical vesselwhen the vessel
thick and its ferther side silvered. The image is formed 23.2 cm behind the silvered face. Find the refractive index ofglass.
is empty. When the vessel is completely filled with a liquid (jo, = 1.5), he observes a mark at the centre B, of the bottom without moving thetelescope or the vessel. What is the height
Solution
of the vessel if the diameter of its cross section is 10cm.
In figure5.141 ABCDht theblock ofglass whose side
is
silvered as shown.
Solution
A
B'
B
In the figure5.140 shown if we consider the position of telescope is at 7" when the point>1 is visible on the rim. Now the same arrangement and position of telescope due to refraction of the ray at point D on water surface.
C
D
when the vessel is filled with a liquid, point B is observed at
C
I*—V—♦p—H
(10.r)
K
20cm
4*— 10cm —*k
23.2cm
H
Figure 5.141
Here in figure we use sin I =
BC
BC
BD
^[(SC)'+(CZ))']
Suppose it appears that the image isformed due tothe reflection at image ofsilvered face 5'C'and ifthe apparent depth ofglass block is X cm then we use
5
20+.v = 23.2 + (10x)
sin I =
.•i:=5.6cm.
Again
ZNDT^ZADC^Zr
sin r =
AC
AC
AD
^l(ACf+(CD)]
As we know the refractive index is the ratio of real depth and apparent depth, we have real depth P'
10 sm r =
Vao'+A')
apparentdepth
10 = — = 1.51.
6.6
# Illustrative Example 5.34
Now using Snell's law, we have
py sin / =
sinr
A light rayfrom air is incident ona glassplate of thickness t and refractive index p at an angle of incidence equal to the
angle of total internal rellection of glass. Compute the displacement ofthe rayduetothis plate in terms ofthickness and refractive index ofglass p. Solution
Figure5.142 shows the ray diagram of the light ray passing through the slab. Here the angle of incidenceis given as '
3cm
5cm
Figure 5.140
sin i = — (As i is equal to critical angle) 4
1Geometrical Optics
237
A J
IrC"v
1 t
I i
r 1
\
«
Figure 5.143
v/
Solution Figure 5.142
BySnell's law,
The image formation of the dust particle is shown in figure5.144
sin;
sinr
= 1^
1
sin/ sm r
^\
In figure wecan seethat displacement of the light ray is given
f
\v'/
as
CEx = BCsin (ir) In figure we also have Figure 5.144
BF cos r =
Here we first consider the image formed byconcave mirror.
BC
For which weuse/? =  40 cmand w=  5 cmfor using in the
t
BC =
mirror formula, whichgives
cos r
I x
sin (Cr)
=
II
cosr
V
_2
t
[sin Ccos rcos C sin r]
40
cosr
= / sin C
, I
sin/ cosC ; X sin C
cos/
R
I_ 5
40
40 v=—cm = 6.67cm
=>
D
As Vis positive, the image F, is formed below the mirror (virtual). The reflected rays are refracted at water surface. The
V(n'+i) 1
depth ofpoint Pj from thewater surface is 6.67 + 5.00= 11.67. Due to presence ofwater, theimage P, will beshifted up and the final position of image is given byitsapparent depth given as
h
Image position = ~ ~
# Illustrative Example 5.35
11.67
=877 cm
Thus final image is formed 8.77cm below the water surface.
A concave mirror ofradius40cm lies on a horizontal table and
water in filled in it upto a height of 5.0cm as shown in figure5.143. Asmali dust particle floats on the water surface at a point P vertically above the point of contact of the mirror with the table. Locate the image of the dust particle as seen from a point directly above it. The retractive index of water is 1.33.
Web Reference at w\v\v.phv$icsgalaxv.coiTi
Age Group  High School Physics  Age 1719 Years SectionOPTICS
Topic Geomciric.il Optics II  Refraction of Light Module Number  1 to 8 and 25 to 32
^ Geometrical
238 Practice Exercise 5.3
opposite to him. He sees thatthereflected andrefracted rays
(i)
comefrom the samepoint which is the centreof the canal. If the 17ft mark and the surveyor's eye are both 6ft above the waterlevel, estimate the widthofthe canal,assuming that the
A converging light beam is incident ontwo glass slabs
of different materials placed in contact with each others (as shown in figure5.145.) Where willthe raysfinally converge?
refractive index ofthe water is 4/3. Zero mark is at the bottom of the canal. [16 feet]
H= 3/2
H= 2
(vi) A ray oflightis incident on a parallelslabof thickness i and refractive index p. If the angle of incidence is 0 then for small 0 show that the lateral displacement of light ray will be Air
r0(pl) 6cm
A =
4cm
— 14cm
Figure 5.145
(vii) How muchwatershould befilled in a container ofheight 21 cmsothat it willappearhalffilled whenviewed alongnormal
[8 cm to the right of second slab]
to water surface. Take refractive index ofwater
(ii)
Find the apparent depth of an object O placed at the
bottom ofa beaker as shown in which two layers oftransparent liquids are filled:
= 4/3.
[12 cm] '
(viii) A glassplatehas a thickness t andrefractive indexp. A light ray is allowed to incident on the plate from air. Findat whatangleofincidence willthe raysrefracted andreflected by the plate be perpendicular to each other ? For this angle of incidence, calculate the lateral displacement of the ray. Kn 1)
h,  25 cm
H, = 1.5
A, = 15 cm
1^2= 2.5
Figure 5.146 [22.67 cm]
(ix) A man standingon the edgeof a swimming pool looks at a stone lying on the bottom. The depth of the swimming pool is equal toh.Atwhat distance from thesurface ofwater is the image of the stone formed if the line of vision makes an angle 0 with the normal to the surface ? H Acos 0
(H^sin^ 0)^'
1
(iii) A point object is placed33 cm from a convex mirror of curvature radius 40 cm. A glass plate of thickness 6 cm and index2.0 is placed between the object andthe mirror, closer to
(x)
the mirror. Find the distance offinal image from the object ?
embedded into the river stands vertically with Im of it above
[42 cm]
the water surface. If the angle of inclination of sun above the horizon is 30°, calculate the length of the post on the bottom
(iv) A light ray falling at 60° angle with the surfece ofa glass slab of thickness Im and is refracted at angle 75° with the surface. Calculate the time taken by the light to cross the slab.
In a river 2m deep, a water level measuring post
surface ofthe river, (p for water = 4/3) [3.44 m]
[xlO"® S]
(xi) A concave mirror of radius of curvature one metre is placed at the bottom of a tank of water. The mirror forms an
(v)
80cm and 40 cm ofthe water in the tank.
A surveyor on one bank of a canal observes the image
image ofthe sun when it is directly overhead. Calculate the distance of the images from the mirror for different depths,
of the 4 inch mark and 17 ft mark on a vertical staff, which is
partially immersed inthewaterand heldagainstthebankdirectly
[47.5 cm, 57.5 cm]
;Geometrical: Optics
.
2391
(xii) A small object is kept at the centre of bottom of a
undeviated attheinterface. Always for image formation we can cylindrical beaker of diameter 6 cm and height 4cm filled consider one rayalong the optical axis and one other rayin completely with water (ti=4/3). Consider thelight ray from an surrounding toproduce image. Infigure5.147(b) we can see all
object leaving the beaker through a comer. Ifthis ray and the paraxial rays from object after refraction meet at the point I rayalong the axis ofbeaker isused tolocate the image, find the which is the image produced due torefraction oflight at the apparent depth in this case.
spherical surfece.
[2.25 cm]
Inabove situation explained infigure5.147, image produced is real image as refracted rays are converging. Depending upon the refractive indices ofthe two media refracted rays may be
5A1 Refraction of Light by Spherical Surfaces
diverging as shown infigure5.148(a) and(b)inwhich thefinal image produced is virtual.
Figure5.147(a) shows a spherical interfece separating two different media 1and 2with refractive indices p, and Here C is the center of curvature of the spherical surface, R is the radius ofcurvature ofthesurface, P isthecentral point onthe
Mediuml
Medium2
(Denser)
(Rarer)
Mi
M2
surface called optic center of the surface and line XX' is
considered asthe Optical Axis' for thegiven setup. Ingeneral the optical axis for the refraction surface can also be called as
principal axis, theterm which we generally usefor spherical mirrors and lenses.
Mediuml
Medium2
(Rarer)
(Denser)
Ml
(a)
M2 Mediuml
Medmm2
„ (Denser)
Object
(Rarer)
(a)
Mediuml
Mediumt2 (Denser)
(Rarer)
(b)
Figure 5.148
Note: Infigure5.148 when object isplaced indenser medium and light rays are refracted toararer medium ftie imageproduced can never be real, it will always be virtual because refracted
rays insuch a case will always bediverging butwhen object is placed inrarer medium asshown in figure5.147 and light rays are refracted to denser medium imageproduced can bereal or (b)
virtual depending upon the refracted rays are converging or
Figure 5.147
diverging. Infigure5.147 refracted rays shown areconverging. Students are advised to thinkand draw ray diagrams for the
If we consider a light ray from an object placed in medium1 situation for virtual image formation in this case on their own incident on the spherical interface at an angle of incidence i carefully. gets refracted atan angle rand therefracted rayinother medium meets theoptical axisatpointI. We canalsoconsider a ray of Similarsituation canalsobeanalysed bystudents for a concave light from the object along the optical axis which passes surfacefeeingthe objectas shown in figure5.149.
Geometrical Optics
240
UsingSnell's law at point Awe have
p, sin / =
Medium2
MediumI
(Rarer)
' ;(Denser)
Forparaxial rayswecan usesmall angles sowe have
p/=p/
I
Object
...(5.45)
I
C
In the geometry of figure shown wecan relatedifferent angles from properties of triangles as ...(5.46) (a)
and
/~0 + a
...(5.47)
Combining above three equations we get Mcdiuni1
Mcdium2
(Rarer)
(Denser)
.p,(0+ a)p2(ct^)
p,0+P2=(l^2~F)aObject '
C
...(5.48)
From tlgure5.148 for small angles we can use the length of normal/JyV/on optical axis as AM= h^i/Q = J?a = v(^ (b)
... (5.49)
If we use coordinate sign convention for this situation as
Figure 5.149
explained in article5.8.8, we can consider P as origin of ft
cooridinate systemall distances toward left as negativeand all Note: In tlgure5.149(b) when object is placed inrarer medium distances toward right as positive so here u will be taken andlight rays arerefracted toa denser medium image produced negative, v and R are taken positive. Sowith sign convention can never be real, it will always be virtual because refracted the angles 0, a and ^ are given byequation(5.49) as rays in such a case will always be diverging whereas in ^ h h , , h tlgure5.149(a) when object is placed in denser medium and 0=—;«=•;: ana u R V lightrays arerefracted to rarermedium image produced canbe realor virtual depending upon the refracted raysare converging Substituting these values in equation(5.48) we get or diverging. In this tlgure5.149(a) refractedrays shown are F2 "1^1 Bconverging. Studentsare advisedto think and drawray diagrams ...(5.50) V n R for virtual image in this case on their own caretully. Above equation(5.50) is called ^Refraction Formula for 5.11.1 Analysisof Image formation by Spherical Surfaces SphericalSurfaces'andit is used for finding theexact location ofimageproduced dueto refraction oflight through a spherical Figure5.150 shows an object placed on the leftof a spherical surface. surface from which a light ray is consideredwhich incidentson the surfaceat pointAand after refraction meetat point/ on the # Illustrative Example 5.36 optical axis as shown.
Mcdiuml
Mediuin2
(Rarer)
(Denser)
Figure5.151 shows a glass sphere of radius 10 cm. Along its diameterfrom one side a parallel beam of paraxial rays incident on it. What should be the refractive index ofglass so
that after refraction all rays will converge at opposite end B. Obicci
R = i Ocm
Fiaure 5.150
Ficure 5.151
IG^eometrlc^t Optics
241
Solution ' 1
I
For refraction formula, we use
V
2x40
10
J__
u =00
^
v=2R = + 2Q cm
1_
V~ 90 20 ~ 80 v=80cm
^=+10cm
Thus final image is seen by observerat a distance 80cm from the poleof curved surface and it is a real image. # Illustrative Example 5.38
Using refraction formula, we have 1^2
A glass rod has spherical ends as shown in figure5.153. The refractive index of glass is p. The object O is at a distance 2R from the surface of larger radius of curvature. The distance
Hi _ H2 Hi R
=>
_H
H1
between apexes ofends is 3i?. Findthedistance ofimage formed
20
10
ofthe point objectfrom right hand vertex. What is the condition to be satisfied ifthe final image obtained is to be real?
.
ji = 2}i—2 r=2
Glass
# Illustrative Example 5.37
r
h—2^—^
3^
Figure 5.153
Figure5.152 showsaglass hemisphere Mofp= —and radius 10 cm. A point object 0 is placed at a distance 20 cm behind the flat face which is viewed by an observer from the curved side.
Solution
Find location of final image after two refractions as seen by
For first refraction at surface A, we use u = 2R'm refraction
observer.
formula as
l^_Hi_ ^ H2H1 v'
u
R,
iL+J__(Hl) v' h = 20cm
2R
...(5.51)
R
Solving for v', we get 2pjg
i? = ID cm
V

Figure 5.152
...(5.52)
(2h3)
For refraction at surface B, we use u = \'3R
Solution
2p^
(2p3)J
m
refraction formula as
After first refraction at flat surface image is produced at a distance given as 3
H/2 = —^20 =30cm 2
1 
For second refraction at spherical surface, for refraction formula
V
+
H2
H]
V
u
H2'Hi
1H {RI2)
H
2\iR
...(5.53)
2p3
we use
w= + 40 cm; i?= + 10cm; ^X•^ =
3
Pj ^
Substituting values in refraction formula, we get _i H2 ~ Hi V
u
R
H= 
3R
2\xR
2p3 Solving equation(5.53) for v, we get
(94p)j? a0n9)(ti2)
Geometrical Optics
1242
The imagewill be real if abovevalue of v is positivefor which finallyrefracted rayswillbe converging, so condition forreal
and
1
cos r =
2 n2
image is
Substituting these values in equation(5.54), we get
(94p)i? (10p9)(^i2)
>0
On simplifying wegetthat it happens when refractive indexof
Sin— =
2
—
R'
1
'F
glass is between 2 and 9/4. Illustrative Example 5,39
p/2
sin— =
p/?
A ray oflight passesthrough a transparent sphereof refractive index p and radius R. Ifb is the distancebetweenthe incident rayanda paralleldiameter ofthe sphere, show thatthe angleof deviation 9 is given by the expression
^
2
ixR' b
=i>
sin —= (/j/p7?^) [(p
sm— =
sm— =
p/?^ [(p2^2_^2)l/2_(^2_£,2y/2]^
~ {R^  b^y^] # Illustrative Example 5.40
Solution
The ray diagram of the light ray passing through the sphere is
A parallel beam of light travelling in water (refractive index = 4/3) is refracted by a spherical air bubble of radius 2 mm,
shown in figure5.154
situated in water. Assuming the light rays to be parallel
(a) Find the positionof the imagedue to refractionat the first surface and the position ofthe final image.
(b) Draw a ray diagram showing the positions of both the images. Solution
Figure 5.154
(a) For refraction at first surface, we use ,
From above figure, the angle of deviation is given as
Mi = (4/3); w= oo andi?j = 2mm .
0 = (/ _ r) + (/  r) = 2/  2r = 2(/  r)
So the position of image due to refraction at first surface is given by refraction formula as Taking sine on both sides of above equation, we get
V
1 Vj
sin— = sin (ir) 2
sin— = sin I cos r  cos i sm r
...(5.54)
Solvingwe get
w
(4/3) 00
1^2i^i R
1(4/3) 2
v, =6mm.
2
Hence the image is formed at a distance of 6mm to the left of
Also from the figure we have
first surface. The ray diagram is shown in figure5.155.
b
sini  ^
cos / =
'F
By Snell's law we have sini smr =
p/?
Figure 5.155
fGebrfieWMr Optics
;243J
(b) For refraction at second surface we use
magnification gives the orientation ofimage whether image formed is on the same side ofoptical axis as that ofobject or
—(6 + 4)= 10 mm,
1^1 ^ ^'^2 ~ (4/3)
^
erected (positive) or itison the other side ofobject or inverted
and R2 =2 mm I (4/3)l
(negative)
V (10) " (2)
,• . . .
• ;
5.11.3 Longitudinal Magnification oflmage
Solving we get V=  5 mm
Figure5.157 shows an object ofwidth dc which is placed on
Sothe final imageis produced at a distance of5 mm to the left optical axis of the refracting surface Sal a. distance x from the
optic center P. After refraction its image / is produced at a
of the second surface.
distance^ from the optic center which is ofwidth dy as shown 5.11.2 LateralMagnification ofImage byRefraction
in figure.
Figure5.156 shows an object ofheight habove the optical axis
ofwhich image produced is ofheight h'. As we have already studied that for areal object ifimage produced isreal, it will be
inverted or on the other side ofprincipal axis. In the figure we have produced image bytwo paraxial light rays, one which is passirig throughcenterofcurvature ofthe interface which will get refracted undeviated and other which is incident on the
optic center ofthe surface at an angle ofincidence iwhich gets refractedat an angle r. Figure 5.l'57
In above figure the distances ofobject and image from the optic center ofthe surface are related by the refraction formula given as
. • 'iiz. _ Hi' = V
.
w
••
.
R
,
Here we use M=—X, =+7? and v=+y which gives if^ ^
^ M'2 "1^1 X
Figure 5.J56
The angles iand r are very small angles as light rays are paraxial
R
^:(fy~dx^O y
h
~h'
~u
V
...(5.57)
Differentiating the above equation we get
so these are related by Snell's law as ...(5.55)

X
. ...(5.58)
From above equation(5.58), we get longitudinal magnification as
Bygeometryin this figure wecan use/= — andA= —so we get
m, =
dx
II
V
Here we iised image height h' and object distance ii negative according to the sign convention.
_ ^V
As already discussed that above relation given by equation(5.59) is only valid for paraxial rays. Here negative sign shows lateraiinversion ofthe image.
5.11.4 Effect ofmotion ofObject or Refracting Surface on
Thus lateral magnification ofimage canbegiven as _
...(5.59)
Image ...(5.56)
Similar tothecase we've discussed incase ofspherical mirrors, here also for small velocities ofthe object orrefracting surfaces, As already discussed that above relation given by the velocity magnification along and perpendicular tothe optical equation(5.56) isonlyvalid for paraxial rays. Thesign oflateral axis can be given by the expressions oflateral and longitudinal h
P2"
GeojTieiri^ljJipt^ magnifications. Figure5.158 shows an object moving with some velocity at a direction shown in such a way that its velocity component along theoptical axis with respect totherefiracting
_ J__J
!_
V ~ 10 20 ~ 20
v=+20cm surface is v^.aiong^^^ direction normal to the optical axis is then the image velocity components can be directly Longitudinal magnification for refraction is given as V
onormal
given as
3l2x{20y ^/•normal
^ ' ^oiwrmal
^ialong
' ^oahng
m
=
m
=
ix{i5y 1.5x400
8
225 Vnnormal
Jmage
V:along ?!
Image size
—xl = r mm.
3
3
^ Vgalong Object V:normal
Lateral magnification m
# Illustrative Example 5.42
A long cylindrical tube containing water is closed by an equiconvex lens offocal length 10 cmin air. Apoint source is placed along the axis of the tube outside it at a distance of
HiV
Longitudinal magnification
, P2"
21 cm from the lens. Locate the final image of the source. Refractive index of the material of the lens =1.5 and that of water = 1.33
Figure 5.158
Glass
Air # Illustrative Example 5.41
^
Water
Source Hi
Figure5.159 shows a smallobject Mof length 1mmwhich lies along a diametrical lineof a glass sphereof radius 10 cm and
h
"21 cm
Figure 5.160
3
p = —which is viewed byan observer asshown. Find the size
Solution
of object as seen by the observer.
We consider first refraction at airglass interface, using refraction formula we get ^ Vi
u
R
Substituting m= 21 cm; w,= 1;/K2 ~ ^ i? = 10 cm
/? = + i?,weget
!:£ J_ _ M
...(5.60)
IX" R
Figure 5.159
The image formed bythe first surface will act as the object for Solution
the second surface and the radius of curvature of the second surface will be taken as  i? for second refiaction as the lens is
For refiaction at glassair interface, we use
u = + 15cm; R= + 10cm; p, = 3/2 and Pj" ^ Substituting values in refiaction formula, we get V
^
M
R
1 3/2 ^13/2 V 15 ~ 10
equiconvex. For glasswater interface, weuserefiaction formula with Wj= 1.5 and^2= 1.33 weget 1.33
1.5
1.331.5
0.17
R
R
...(5.61)
Adding equations(5.60) and (5.61), we get 1.33
1
V "^21
0.67
...(5.62)
i'Geometrical 6'ptics 245
The focal length ofthe lens in air is 10 cm. So by using lens sin
maker's formula we have 1
^1 = (1.51) 1)
10 or
^(1cos^r) 1/2
I
1^
7!2
'y 1 .7!'^7!
73x0.1 I
M 2R
i?=2.5cm
4R
^ = ——7(43) =0.086 metre Total height oflight ray above the plane mirror is
...(5.63)
Substituting the value of^ in equation(5.62) and solving it, we get
H= 7? + /i = 0.1 + 0.086 = 0.186 metre From figure we have X = 7? cot /
V = 70 cm
Thusfinal image is formed 70cminside thetube andit isreal.
also we use
(7 = 7! +a:= 7?(1 + coti)
=>
(7=7!(l+cot2r) =7!ri +^^l V s\n2r)
# Illustrative Example 5.43
2cos^ /•!
(7=7! 1+2 sin r cosr
A cylindrical glassrod of radius0.1 m and refractive index
/
lies on horizontal plane mirror. Ahorizontal ray oflight moving perpendicular to the axis of the rod is incident on it. At what
uV(H?74)
leaves the rod at aheight of0.1 mabove the plane mirror? At what distance a second similar rod, parallel to the first, be placed on the mirror, such that the emergent ray from the second rod is in linewith the incident rayonthe firstrod ?
(2n^/4)]
(7= 7! 1 +
height from the mirror should the ray be incident so that it d= R 1 +
1/2
75/2
= 0.1
l+j= 75
.
Distance of the second rod from the first rod is
Solution
2(7 =
.
75
= 0.3155 m.
The situation and raydiagram isshown infigure5.161. #Illustrative Example 5.44
\
\
I I'' / H
\
v ^
Aquarter cylinder ofradius Rand refractive index 1.5 is placed
\ fl
on a table. Apoint object P is kept at a distance mR from it.
To
Find the value ofwfor which aray from Pwill emerge parallel to the table as shown in figure5.162.
I*/?
'
i«—^
Id
4
Figure 5.161
From figure we have i = 2r and we have
h
\*—mR
AO sin i
Figure 5.162 Solution
= R sin i
= 7?sin 2r=2R sin rcosr By Snail's law we have
...(5.64)
To solve this problem, we use the principle ofreversibility of light.
sini
p. 
H
sin2r
—
= 2 cos r
To analyze the situation we consider that the light isincident
sinr
from therightparallel tothehorizontal surface oftable asshown
From equation(5.64) we have
and we find where it converges on the table after emerging from the plane surface. This must be point P ifpath oflight ray
h2R sin r cosr
is reversed as specified in the question.
Geornetric^;jOpticsj
246
diameter ofthesphere from the sideon which it lies. How far fromthe surfacewill it appear.Refractiveindex ofglass is 1.5. [2.5 cm behind the surface] •
C
' K
B
\A
(iii) Arayoflight refracted through a sphere, whose material has a refractive index p in such a waythat it passes through
♦K R H
mR
Figure 5.163
the extremities oftwo radii which make an angle 0 with each other. Prove that if a is the deviation of the ray caused by its passage through sphere
Usingrefraction formula for curved surface, \ye have V
u
R
cos(0  a) = p cos 0/2
According to the given problem we use
u=oo;
1,5; Pj = 1and^ = +i?
(iv) A ray is incident on a glass sphere as shown in figure5.165. The opposite surface of the sphere is partially
Substitutingthe values in refraction formula, we get 1.5
1
1.51
V
CO
R
:=>
silvered.If the net deviationofthe ray transmittedat the partially silvered surface is 1/3^ ofthe net deviation suffered by the ray,
,
V=3R ^
When the lightrayincident ontheplane surfece, it gets refracted and produces an imageat a distance 2R/ii= 2R/1.5 = 4R/3.
reflected at the partiallysilvered surface (afteremerging outof the sphere), findthe refractive indexof the sphere. Partially silvered
From the given condition ofwe can use 4R mR= —
4m =
Figure 5.165
Web Reference at www.phvsicsgalaxv.com
[S ]
Age Group  High School Physics ] Age 1719 Years Section  OPTICS
(v)
Topic Geometrical Optics II Refraction of Light •
near normal incidence. Show that the image in terms of the
A parallel incidentbeam fallson a solidglass sphereat
Module Number 16 to 24 and 33
index of refraction p and the sphereof radiusR is givenby 2(pl)
Practice Exercise 5.4
(vi) (1) A spherical surface S separates two media I and 2 as shown in figure5.164. Find where an object 0 is placed in mediumI so that the light rays from object after refraction becomes parallel to optic axis of this system. •s
h = 4/3
/
.
^
~
H2 = 3/2
A hollow sphere of glass of refractive index p, has a
small mark on it interior surface which is observed from point
outsidethe sphere on the side opposite the centre. The inner cavity is concentric withthe external surface andthe thickness of glass is everywhere equalto the radius ofthe inner surface. Prove that the mark will appear nearer than it really is by a
distance(p  1)i?/(3p 1), whereR is the radius of the inner surface.
u  240 cm Mark
Figure 5.164 [ 240 cm]
(li)
A glass sphere of radius 5cm has a small bubble at a
distance 2cm from its centre. The bubble is viewed along a
iR
Figure 5.166
^eometriparpptics' "' •247
(vii) Atransparent sphere ofradius Rhas acavity ofradius R
2 as shown in figure5.167. Find the refractive index ofthe Fish
sphere ifaparallel beam oflight falling on left surface focuses at point P.
Figure 5.169
[3.84 mm/s]
(xii) Figure5.170 shows an irregular block of material of Figure 5.167
refractive index ^/2. Arayoflight strikes the face AB as shown in figure. After refraction itis incident on aspherical surface CD ofradius ofcurvature 0.4m and enter amedium ofrefractive
index 1.514 to meet PQ at E. Find the distance CDupto two places ofdecimal.
(viil) Alight ray incident at apoint on the surface ofaglass sphere ofp = Vs atan angle ofincidence 60°. Itisreflected and refracted at the farther surface of sphere. Find the angle between reflected and refracted rayat this surface. [90°] u= i.514
(ix)
E
Q
Aglass sphere (p = 1.5) with a radius of15.0 cmhasa
tiny air bubble 5 cm above its centre. The sphere is viewed looking down along the extended radius containing the bubble. What isthe apparent depth ofthebubble below thesurfece of the sphere ?
Figure 5.170
[6.06 m]
[8.57 cm from top]
(x) Figure5.168 shows aglass hemisphere placed on awhite horizontal sheet. Avertical paraxial light beam ofdiameter d incident on the curved surface ofhemisphere asshown. Find 5.12 Total Internal Reflection thediameter ofthelight spot formed onsheet after refraction.
Figure5.171(a) shows alight ray refracting through an interface oftwo media. The incident light ray is travelling in denser medium and after refraction itbends a\yay from normal as we know and this figure also shows the reflected ray which is always present
in all cases ofincidence and it is reflected at the same angle of white sheet
Figure 5.168
incidence. If we increase the angle of incidence as shown in figure5.I7I(b), theangle ofrefraction also increases. As the angle ofrefraction is more than the angle of incidence it will
approach to 90° at some specific value ofias shown. This angle ofincidence atwhich the refracted ray grazes along the interfece ofthe two media is called 'Critical Angle' for this pair oftwo media. Figure5.171 (c) shows alight ray incident on the medium
(xi)
Figure shows afish bowl ofradius 10 cm in which along boundaryatan angle slightlygreater than critical angle at which
a diametrical line a fish Fis moving atspeed 2mm/sec. Find the no refracted rayexist and only reflected ray exist. This iscalled
speed offish as observed by an observer from outside along
Total Internal Reflection' and it occurs when a light ray
same line when fish is at a distance 5 cm from the centre of
travelling in adenser medium incident on a boundary ofararer medium atan angle greater than the critical angle.
bowl torightofit as shown in figure5.169.
Geor^trical O^icsJ^ 1248
5.12.1 Refraction ofLightRaysfrom a Source in a Denser Medium to Air
Figure5.172(a) shows apoint isotropic source oflight placed inside water at a depth h below the airwater boundary. The Rarer Medium (i,)
light rays which incident close to normal as shown will get
• Denser" Medium (1^2)
refracted toairand diverge. Ifwe see aspecific light ray which isincident ontheairwater interface at critical angle will graze
along the interfece and all the rays from source which incident atangle greater than this angle will get totally reflected into water. In this way we can say that light rays from source will emerge out in air only from a circular region as shown by
(a)
perspectiveview in figure5.172(b). Here we can see that all the light rays from the source on the surface ofaconical region of halfangle 0^ will graze radiallyalong the airwater interface and all light rays from source felling on interface within this conical region will escape to air and all the light rays falling on interface Grazing Refraction
outside this conical region will get totally internally reflected and will not come out in air.
•'
>
r = Atan 6,
water
(b)
Source
Rarer Medium (u.)
(a)
/•= Atan 9r.
Denser Medium (u,)
(c) , Figure 5.171 •"
Source
The critical angle can be given bySnell's law attheboundary as P2 sin 0c
(b)
Figure 5.172 sin U.= ^2
Infigure5.172(a) inDSOA wecanwrite
0 =sin'l — >2
For a light ray passing through a denser medium having
r = htan 0,
...(5.65)
refractive index pincident ontheboundary ofair(^3;^= 1) inthe surrounding ofthe medium then critical angle for this medium Equation(5.65) gives the radius ofthe circle ontheinterface just above the point source oflight from which light rays escape can be given as to air. Byusing the concept ofsolid angle ofthis cone wecan also find the fraction of total light power of the source which
gets refracted to airin this case. The solid angle ofthe cone
IGeqmetrrcai'Optics
shown infigure5.172(b) with halfangle 6^ is given as n = 2;i(lcos0^)
=>
n2jt(l^lsin^0^ ) r
=>
'
When a light ray incident on the boundary oftwo media from the side of denser medium in a grazing manner at incidence angle 90° as shown in figure5.174 then the light ray passes as it is incident on the boundary as the incidence angle is more than critical angle so it gets intemally reflected.
fl = 27r 1 l'—r (As sincos0= —)
 ' I V It' J
Rarer Medium
At the source all light rays are emitted isotropically in total solid angle An steradian in which total power of light source is distributed uniformly. If we consider the source power is P watt then the light power which is escaping out from water to air is the power within the shaded conical region as shown in figure5.172(b) which is given as
Denser Medium
Figure 5.174
5.12.3 Refraction by a Transparent Medium of varying
•^to air
^^
Refractive Index
,„±
The fraction of light power of source which escapesfrom water
When a light ray incident on the surface of a medium having continuously varying refractive index with distance then the path of light ray is not a straight line. Figure5.175 shows a
is given as
medium in which we consider that its refractive index increases
^ ^toair 4^ ^
/=
"7
5.12.2 Cases of Grazing Incidence of Light on a Media Interface
We know that path of a light ray in cases of reflection and refraction is retractable and we have studied that a light ray
travelling in denser medium when incident on a boundary of rarer medium at critical angle will graze along the boundary as shown.in figure5.173(a). So by the concept of reversibility of light we can state that a light ray travelling in a rarer medium when incident on the boundary ofa denser medium at angle of
with xcoordinate of the position. To understand the path of light shown in this figure we consider an elemental slab of medium at a positionx and having thicknessdx which is shown in figure by a shaded region. When a light ray incident on this slab at an angle 0 then it will slightly bent toward normal as refractive index ofthe medium slightly increases on the right side ofslab. In the same sense as continuously refractive index of the medium is increasing to the right, the light will continuouslybend toward normal to each elemental slab which is resulting in a curved path as shown in the figure.
Glass slab having
incidence approximately 90°whichis calledgrazing incidence of light asshown in figure5.173(b), the light will enter in the denser medium at critical angle0^.
where/is an increasing function
Rarer Medium
Grazing Refracted Ray
: . Denser Medium
Figure 5.175
In above figure, insidethe mediumfor all parallel normals,we
can relate theangles cfjj and (jjj by Snell's law at points AandB where refractive indices are pj and pj as given below.
(a)
Grazing Incidence
j.] sin(90°(j),)= ^2sin(90°(l)2)
Rarer Medium
Denser Medium
=>
jUj cos4>, = }i2 cos 42 =>
# Illustrative Example 5.48
1 sin i > — V
From figure, we have
A rod made of glass (p = 1.5) and of square crosssection is bent into the shape shown in figure5.182. A parallel beam of light falls perpendicularly on the plane flat surfece^f. Referring to the diagram, d is the width ofa side and R the radius ofinner semicircle. Find the maximum value of ratio {d/R) so that all
sm I ~
PO
R
OQ
(R + d)
R
1
(R + d)
1.5
iGeometrrcalJDptics
:253:
l.5R>{R + d) 0.5R>d
or
Substituting the values of a and weget d/R
f=sin' [ff sin {45°sinVw,/ff)}]
178° clockwise
(ii) For the ray to pass through the diagonal face undeviated, theraymust strike normally on theface, i.e..
#Jllustrative Example 5.55
A right angle prism (45° 90° 45°)ofrefractive index n hasa
sin I
plate ofrefractive index w,(«, < w) cemented to its diagonal face, The assembly is in air. Aray is incident on AB (figure5.201)
sin 45°
= 1.352
or
sin / = sin45° ^ (1.352)
or
sin I
1.352
A
= 0.9562
/ = sin' (0.9562) = 72.9°

/
# Illustrative Example 5.56 n
B
In figure5.203, light refracts from material 1into athin layer of
Figure 5.201
material2, crossesthat layer, and then is incidentat the critical
(i) Calculate the angle ofincidence ziAB for which the ray
angle on the interface between materials 2 and 3.
strikes thediagonal face at thecritical angle. (ii) Assuming«= 1.352,calculatetheangleofincidenceat/45
for which the refracted ray passes through the.diagonal face undeviated. «,= 1.60
Solution Figure 5.203
(i) In the given situation we consider the critical angle for face ACbe0^ Then bySnell'slawwehave sinQ^= {n^/n)
(a) What is the angle 0 ? ...(5.93)
Let the required angle ofincidence be / and angle ofrefraction at faceAB is r asshown in figure5.202. From figure, or
...(5.94)
p, sin/, = i, sin/2 = ^^sin 0^, Forinterfaceof medium 1and 2, we use
Forrefraction at/'(using Snell's law), wehave sin / = « sin r
Solution
(a) By Snell's law we have for the two interfaces
r+0^=45°
r = (45°0^)
(b) If0isdecreased, isthere refraction oflight into material 3?
...(5.95)
(1.6) sine = (1.80)
1.30 1.80
,(\3) (b) If 0 is decreased, then angle of incidence on the second
interfece will decrease from the critical angle. So light refraction will occur into medium3. Figure 5.202
Geometrical bpticP
[260_^ , # Illustrative Example 5.57
Socoordinates of image formed afterreflection from concave
Findthe coordinates ofimage ofthe pointobject 'O' formed
mirrorare fr^175cm,—cm [.
after reflection from concave mirror as shown in figure5.204
assuming prism to be thin and small in size and ofprism angle 2°. Refractive indexofthe prismmaterialis 3/2. /= 30 cm
'X
(0,0)'
Monochromatic light is incident on a plane interface AB between two media of refractive indices «, and «2 («2 ^ "i)
anangle ofincidence 0as shown infigure5.206. The angle 0 is infinitesimallygreater than thecritical angle for thetwo media so that total internal reflectiontakes place.Now if a transparent slab DEFG of uniform thickness and refractive index is introducedon the interface(as shown in figure), show that for
•20cm
K5cmf
Illustrative Example 5.58
Figure 5.204
any value ofWj all light will ultimately be reflected back again intomedium II. Consider twocases (i)«3 < «, and (ii)n2>ny
Solution
Medium III («,) D\
When light rays passes through prism, allincident rays will be
"
1 1
deviated by the deviation angle given as
Medium III ("3)
i
Si
^2
1
IF
G\
J Medium II (wj)
5=(^lM=(fl]2° =r=^rad
Si
Asprism isthin,object and image will beinthesame plane as
Figure 5.206
shown in figure5.205 below. Solution
(i) Inthis case asW3«,wehave
« «2 if"we consider 0^. be the critical angle for medium and ^2 thstiwe have sin 0=
Figure 5.205
Thus in this casethe shift of image due to prism in ydirection is given as
Shift^=a5=5[y^Q^
It is given that 0 is infinitesimally greater than 0^^. Here we
consider refraction from «2
where
< ^2 ^itd if0^^' be the
critical angle for this interface,then we have sin 0'= —
c
W2
With the given conditions wehave 0^^' < O^ and so 0 > 0(.'. Thus the raywillbe totallyreflected backintothe medium n^
Now thisimage will actas an object for concave mirror. Sofor
(ii) In this case when ^2 ^ "r fti this case we have
mirror formula we use
M=  25 cm and/=  30 cm uf
= 150 cm
v =
sin Qr'~ — «2
Butasperthegiven conditions we have 0^'> 0^^. and 0< 0^'so theray incident at 0, willberefracted intomedium ^3. If 0' be the angle of refraction,then by Snell's law, we use
Also,
m= "
= + 6
sin0'
u
Thus distance of image from principal axis is given as n TZ
^
W3
n
——x6 = — cm 36
sin 0
6
sin0'=sin0x
...(5.96)
{Geometrical Opt^
261
As 0 isslightly greater than G^we can use , Web Referenceat www.physicsgalaxv.com sin0 =
; Age Group High School Physics  Age 1719,Years «2
SectionOPTICS
Topic Geometrical Optics 11 Refraction of Light
Thus fromequation(5.96), wehave
; Module Number  9 to 15 and 34 to 44
=>
n,
rio
n,
«2
"3
«3
sin0'=—X ^=!
Thus the light ray refracted into
will be incident at the interfece
DE at an angle 0' which is slightly greater than the
corresponding critical angle. Thus the ray will be totally
reflected back into
and finally transmitted to n^.
Practice Exercise 5.5
(i)
A glass prism ofangle 72° and indexof reflection 1.66is
immersed in a liquid ofrefractive index 1.33. Find theangle of minimum deviation for aparallel beam in light passing through the prism.
# Illustrative Example 5.59
[22° 22 T
Aprism with prismangle 60° and refractive index JTfl isgiven.
(11)
In figure5.208 shownfind the angle of incidence ofthe
Findtheminimum possible angle ofincidence, sothatthelight light ray on faceABof the prism forwhichlight willreach face ray is refracted from the second surface. Also find 8
. max
AC at incidence angle 60°. /I = 90°
Solution
Asweknow thatminimum andmaximum incidence angles are corresponding tothe maximum deviation of light raythrough the prism. Figure5.207showsthe ray diagram for the case of minimum incidence angle.
Figure 5.208 [90'
(111)
A ray of light is incident on a glass slab at grazing,
incidence. The refractive index of the material of the slab is
given by p,= s/Cl +y) .Ifthe thickness ofthe slab is d, determine the equation of the trajectoryof the ray inside the slab and the coordinates of the point where the ray exits from the slab. Take Figure 5.207'
the origin to be at the point of entry of the ray.
According to Snell's law, we have b= vl
l^sin/^j„= Jj sin(AQ A point source of light is placed directly below the
sm i^.^= ^jj (sinA cos CcosA sin Q sin lmin=^/J
sin60JlycosbOJy
surface ofa lake at a distance h from the surface. Find the area
on water from which the light will come out from water. TZ/l
(M 1)
L. =30'
(v) A ray of light incident normally on one face of the faces ofa right angledisosceles prism is foundto be totallyreflected.
6^ = i^^ + 90°A = 30°+ 90°60° = 60°.
What is the minimiun value ofthe refractive index ofthe maierial
mm
Geometrical OpticsJ
262
of the prism? When the prism is immersed in water,trace the path of the emergentraysfor the same incident ray, indicating values ofemergence angle,
= 4/3)
[1.414, 48°35']
(vi) Figure5.209 shows a triangular prism ofrefracting angle 90°. A ray oflight incident at face ABat an angle 0 refracts at point 2 with an angle of refraction 90°. (a) What is therefractive indexof the prism in termsof 0 ? (b)What is the maximumvalue that the refractive index can have ? What happens to the light at O ifthe incident angle at Q is (c) increased slightly and (d) decreased slightly?
(ix) A light ray composed of two monochromatic components passes through a trihedral prism with refracting angle A = 60°. Find the angle AS between the components of the ray after it passes through the prism if their respective indices ofrefraction are equal to 1.515 and 1.520. The prism is oriented to provide the least deflection angle. [0.5° approx]
(x) A parallel beam of light falls normally on the first faceof a prism ofsmall angle.Atthe secondface, it is partlytransmitted and partly reflected.The reflected beam strikes at the first face again, and emergesfromit in a directionmaking ao angle 6°30' with the reversed direction ofthe incident beam. The refracted
beam is found to have undergone a deviation of 1°15' from the original direction. Find the refractive index ofthe glass and the angleoftheprism.
Figure 5.209
[(a) n/i +sin"0 ; (b) V2 ; (c) No grazing emergence; (d) Total Internal ReOection]
(vil)
On one face of an equilateral prism a light ray strikes 3
normally. lfitsp= —, find the angle between incident ray and the ray that leaves the prism. [60°]
(viii) A long rectangular slab of transparent medium of thickness d is placed on a table with length parallel to xaxis and width parallel to the^axis. A ray oflight is traveling along j'axis at origin. The refractive index p ofthe medium varies as
p= Vr+^^ .The refractive index ofthe air is 1.
(xi) A ray oflight is incident at an angle of60° on one face of a prism which has an angle of30°. The ray emerging out ofthe prism makes an angle of30° with the incident ray. Show that the emergent ray is perpendicular to the face through which it emerges and calculate the refractive index of the material of prism. [73 ]
(xii) The refracting angle of a glass prism is 30°. A ray is incident onto one of the faces perpendicular to it. Find the angle 5 between the incident ray and the ray that leaves the prism. The refractive index ofglass isn= 1.5. [18.6°]
(xiii) A ray oflight travelling in air is incident at grazing angle (incident angle = 90°) on a large rectangular slab ofa transparent medium ofthickness t = 1.Om(see figure5.211). The point of incident is the origin A(0,0).
(a) Determine thexcoordinate ofthe pointy4, where the ray intersects the upper surface of the slabair boundary. (b) Write down the refractive index ofthe medium at.^.
Air
1= I.Om
Ax,.v,)
/(x,y)
Glass slab Air
A{0. 0)
Figure 5.211 Figure 5.210
[2d In (2), VJ 1
The medium has a variable index ofrefraction fi(y) given by
l)'/2
:^Ge6metricai Optics where A: = 1.0 m"^^.The refractive index of air is 1.0.
Convex and Concave lenses shown in figure5.212 are basically (a) Obtain a relation between theslope ofthetrajectory ofthe lens families and these are further categorized in three ways. rayat point 5(x,y) in themedium andtheincident angle atthat Convex lenses are of three typesas given below point.
(b) Obtain an equation for the trajectoryy{x) ofthe rayin the medium.
(c) Determine the coordinates (x,, y,) ofthe point P, where the rayintersects the upper surface ofthe slabair boundary. (d) Indicate the path of the ray subsequently.
1. Biconvex Lenses : Theseare the lenses bounded bytwo intersecting sphericalsurfaces having their center ofcurvature on different sidesofthe lensas shown in figure5.213.
2. Plane Convex Lenses : These arethe lenses bounded by one spherical surface intersecting with one plane surface as shown in figure5.213.
[(a) e = coti
•Ah)x4k = 47""; (c)4, 1; (d)
"li
3.
Concavo Convex Lenses: These are the lenses bounded
by two intersecting spherical surfaces having their center of curvatureon same side of the lens as shownin figure5.213. 5,
5.14 Thin Lenses
Lensesare the transparent medium bounded bytwo spherical surfaces. There are two types of lenses in general based on thickness ofthe lens at its center, thin and thick. Both lenses
differ in context of image formation and analysis of image characteristics. One byonewewill discuss upon both butmore
Biconvex lens
important is to understand ^Thin Lenses^ which are most commonly used lenses in practice.
Figure5.212 shows the wayhowlenses are madebyusing two spherical surfaces. Figure5.212(a) shows a case when the two
surfaces bounding a region in between intersect at the edges and figufe5.212(b) shows the case when the two surfaces
bounding a region in between do not intersect at the edges. First one is called ^Convex Lens^ and second one is called 'Concave Lens'.
Piano convex lens
Concavo convex lens
S,
Figure 5.213
In the manner described above for convex lenses, concave lenses are of three types as given below
Convex lens
1. Biconcave Lenses: These arethe lenses bounded bytwo nonintersecting spherical surfaces having their center of
(a)
curvatureon differentsidesofthe lens as shown in figure5.214.
2. Plane Convex Lenses ; These are the lenses bounded by one spherical surfece not intersecting with one plane surface as shown in figure5.214. 3. Concave lens
(b) Figure 5.212
Concavo Convex Lenses: These are the lenses bounded
bytwo nonintersectingsphericalsurfaceshaving their center of curvatureon same sideof the lens as shownin figure5.214.
Geometrical Opticsj
Due to the above behaviour all convex lenses placed in rarer
surrounding medium are called ConvergingLenses\^^ all concavelensesplaced in rarer surroundingmediumare called 'Diverging Lenses'.As shown in figure5.215 wecan seethat the converging lenses have a 'Real Focus' on the side oflens where refracting raysexistanddiverging lenses havea' Virtual Focus' on the side ofthe lens where incident ra>^ exist.
Biconcave lens
The behaviour of both type oflens frmilies get interchanged when the surrounding is denser than lens material. This is explainedin figure5.216(a) and (b)
Piano concave lens
Convexo Concave lens
Figure 5.214
5.14.1 Converging and Diver^ng Behaviour of Lenses F'
In general lenses having made up of material denser then their surrounding like most common is the uses ofglass lenses in air. In such cases when a parallel beam of light incident on the lenses, all convex lenses converges the beam after refraction to a point and all concave lenses diverges the beam after refraction which appear to come from a point behind the lens as shown in figure5.215(a) and (b).
•
Focal point (focus) of convex lenses in denser medium (a)
/V 4'
\/
Focal point (focus) oflens
oflens
' Focal point (focus) oflens
(a)
Focal point (focus) of concave lenses in denser medium
Figure 5.216
Note: Behaviour oflenses  converging or diverging is defined only for parallel incident rays on it, there should not be any misconception developed that a converging lens always Focal point (focus) oflens
Focal point (focus) oflens
converges all type of light rays. Its defined for parallel rays which a converging lens converges and a diverging lens diverges when the surrounding medium on the two sides of lens is same.
5.14.2 Priiuary and Secondary Focus of a Lens Focal point (focus).of lens (b) Figure 5.215
When light rays are refracted from a lens then there are two focal points defined for a lens, primary and secondary for both concave and convex lenses. Figure5.217(a) shows primary
fGeometrical Optics

'26^
focus of a convex lens. This is the point on one side of lens
where ifan object is placed then after refraction all light rays
5.14.3 Standard Reflected Light Rays forImage Formation by Thin Lenses
will become parallel to the principal axis of the lens. Figure5.217(b) shows the primary focus of a concave lens. There are three standard incident paraxial rays and their
This isthe point toward which ifa converging beam oflight is corresponding refracted rays after double refraction through a incident on lens then after refraction these light rays become thin lens, using which we can roughly analyze thelocation of parallel to principal axis of the lens. image on the principal axis ofthe lens. Any two ofthe three standard rays we can use for finding the relative position of
image produced and understanding ofthese rays also helps in
Real
object
analyzing the ray diagram for image formation by a lens in different situations. We will discuss these rays one by one.
Ray1: Incident onthelens parallel totheprincipal axis Figure5.219(a) and(b) shows aparaxial light rayincident on theconvex and concave lens which isparallel toprincipal axis which after refraction from the two surfaces ofthe lens passes
(a)
through the primary secondary focus of the lens. In case of Virtual object
(b)
convex lens itactually passes through thefocal point whereas in case of concave lens it appears to pass through the focal point as shown.
.
Figure 5.217
Figure5.218(a) shows parallellight raysincident ona convex lens which afterrefraction throughthe lens meetat the focus on the other side of lens, this is called secondary focus ofthe convex lens. Similarly when parallel raysincident ona concave
lens as shown in figure5.218(b), after refraction light rays (
(
diverge in such a waythatthese appear tocome from the focal II. point located behind thelens, thisiscalled secondary focus of (
I,
It
II
s
the concave lens.
1
(b) Figure 5.219
Ray2: Incidentonthelens directly at theopticcenterofthe Real Image /
H
(a)
lens
Figure5.220(a) and(b) shows aparaxial light rayincident on theconvex andconcave lens at the optic center ofthe lens. As lensis thinatthecenter it behaves like a glass slabas shown in insetview so the light ray passesundeviatedwith small lateral
^
"
displacement which can be ignored and considered the ray passing straight if incident on the optic center of the lens.
virtual Image
(b)
Figure 5.218
Note : For thin lenses the focal length of the lens for both
primary and secondary focus is same. This we can also prove afterunderstanding the calculation offocal lengthofthe lensin coming sections. (a)
Geometrical Optics J
•266
CaseI: Obj ect Islocated at infinity
Asalready discussed while analyzing reflection cases that there are two possibilities of an object at infinity. One is when all incident rays areparallel toprincipal axiswhen image produced is real and located at focus of the mirror as shown in
figure5.222(a) andother possibility iswhen incident rays areat someangleto principal axiswhen image produced is real and located in focal plane as shown in figure5.222(b). In both of these cases we have considered the image produced is highly diminished in size, real and inverted (produced on the other side of principal axis as that of object).
(b)
Figure 5.220
Ray3 : Incident on the lens along the direction of primary focus
Figure5.221(a) shows a paraxial raypassing through theprimary
2F
focus ofthe convex lens arid then incident on the lens, this ray after refraction becomes parallel to the principalaxis ofthe lens.
Similarlyfora concave lens,a light ray which isincidenton the lens along the linepassing through its primaryfocus as shown in figure5.221 (b)becomes parallel after refraction.
Focal
plane
(b)
Figure 5.222
Casen: Object is located beyond 2F point F,
Figure5.223 shows this situation in which object is a small candle and we find the image oftip of the candle by considering
(b)
Figure 5.221
ray1 and ray2 as discussed in article5.12.3. With this ray diagram we can conclude that the image produced for this location ofobject(placed beyond 2Fpoint on principal axis)is
Now using these casesof paraxial incident and refracted rays
real, located between F and 2F, inverted (on other side of
we can discuss various cases of image formation by convex and concave mirrors for different positions ofobject.
principalaxis)and smallerin size than that ofobject.
5.14.4 Image Formation by Convex Lenses
Whenever an object location is given for a convex lens then using any two of the three incident rays and corresponding refracted rays discussed in article5.12.3 vye can find the image location using ray diagram. There are some positions near to principal axis in different regions where if an object is placed, using ray diagram we can get some informationabout the image formation. This information is very helpful in rough analysis while solving different types of questions based on image formation. For both convex and concave lenses we are going to discuss different cases for position of object in front of the lenses and its corresponding image produced. First we will take up the cases for convex lenses in which there are five possible cases we will discuss for a real object in front of the convex lens.
Figure 5.223
Casem: Object is located at 2F point on Principal Axis
Figure5.224 shows this situation in which object (candle) is located at 2F point and we find the image of the candle by considering ray1 andray2 as discussed in article5.12.3. With this ray diagram we can see and conclude that the image produced for this location of object (placed at 2F point on principal axis) is real, located at 2F point on other side of lens, inverted (on other side ofprincipal axis) and ofsame size as that of object.
^Geometrical Optics
267
produced is located behind the lens, virtual, erected (on the same side ofprincipal axis) andenlarged. IF 2F
F
\1 Figure 5.224 f >
CaseIV : Object is located between F and 2F points on
11 X
. , li 2F
f
11 X
2F
•.s
F
KT
Principal Axis
Figure5.225 shows this situation in which object (candle) is located between Fand 2Fpoints and we find the image ofthe candle by considering ray1 and ray2 as discussed in article5.12.3. With this ray diagram wecan seeand conclude
Figure 5.227
Above sixcases aregeneral cases which explains therelative position, nature and orientation ofimage produced byaconvex
that the image produced for this location of object (placed between Fand2Fpoints on principal axis) isreal, located beyond 2Fpoint, inverted (on other side ofprincipal axis) and enlarged
lens for a real object placed close to its principal axis using
compared to that of object.
give an idea about the estimation ofimage produced upto some
paraxial rays. While solving different questions above cases extent before going for theexact process of image formation.
IF
2F
F
Figure 5.225
CaseV: Object is located at Focus
5.14.5 Image formation by Concave Lenses
Unlike to the different cases of image formation bya convex lens for different positions of object in front of it, in caseof a concave lens there areonlytwo possibilities ofimage formation for a real object placed in front of it.
CaseI: Objectlslocatedatinfinity
Figure5.226 shows this situation in which object (candle) is located at focal point ofthelens and we find the image ofthe When light rays from distant object fall on a concave lens,
these rays diverge afterrefraction in such awaythat for incident candle by considering ray1 and ray2 as discussed in rays parallel toprincipal axis a image is obtained at focal point article5.12.3. Withthis ray diagramwe can seethat the refracted ofmirror as shown in figure5.228(a) andfor incident rays non raysare paralleland willproduce imageat infinity. Sowecan
conclude with this diagram that image produced will be atinfinity, inverted (on the other side ofprincipal axis) arid highly enlarged.
parallel to principal axis image is obtained in focal plane as shown infigure5.228(b). The image produced will bevirtual,
diminished and erected (produced on the same side ofprincipal axis where object is located).
Figure 5.226
CaseVI; Objectislocated between Focus and OpticCenter
Figure5.227 shows this situation in which object (candle) is located between O and Fand we findthe imageof the candleby considering ray1 and ray2 as discussed in article5.12.3. With
this raydiagram wecan seethat theserefracted raysfrom lens arediverging in a waythat theseappearto becomingfrom the point/behind the lens from a virtual image. So forthis location of object (placed between O and F on principal axis) image
(b).
Figure 5.228
Geometrical Optics 1,
268
Again with same sign convention we can use « = v =/ Figure5.229 shows a situation in which theobject (candle) is i, = )x,li2=l and ^ =^2 and we get placed infront oftheconcave lens and tofind image using ray = ...(5.98) diagram we consider ray1, ray2 and ray3 as mentioned in / V, F,
CaseII: Object is placed anywhere in front ofthe lens
article5.12.3. Wecan seein this figure that the imageis formed
byback extension ofthereflected rays corresponding tothese incident rays from theobject and theimage isproduced behind the mirror betweenFand (9, virtual, diminished and erected(on the same side of principal axis as that of object).
Note : In above equations(5.97) and (5.98) we have not
considered the signs of u, and R^ as the lens can be of different types among thesixtypes asexplained in article5.12. Adding the two equations, we get 1
1
R
R•2j
...(5.99)
Above equation(5.99) is called 'Lens Makers Formula'' used to calculate the focal lengthofa thin lens.With this expression we can see that focal lens of thin lens for both primary and
secondary focus remain same on thetwo sides ofthelens. The
Figure 5.229
aboveformula is usedwhen a thin lensis kept in air and ifa thin lensofrefractive indexp. iskqjt in a surrounding medium having
5.14.6 Focal length of a thin lens
refractive index
then the above formula can be modified as
Figure5.230 shows a thin convex lens onwhich parallel light givenbelow in equation(5.99). ra>^ incident from leftandaftertworefractions at itsspherical 1 ( II
surfaces ofradius i?j and
light rays converge to its focal
1^ R2 ,
/
point at adistance called the^Focal Length' ofthelens and it is denoted by. Wewill findoutthis focal length/by usingthe analysis of refraction of light at the twosurface.
\
(sioo)
5.14.7 Focal length ofdifferent types ofstandard thin lenses
Byusing Lens Makers Formula wecanfindthefocal length of any thin lens but for quick reference we are giving here the
52(^2)
directrelationsfor magnitudeof focallength of somestandard shaped lenses.
' .
1. Equiconvex or Equiconcave Lenses : These lenses have same radiiofcurvature ofthetwo surfeces soweusei?j = ^2 F andforthisequation(5.99) willreduce tothe form given below
Figure 5.230
because the two center ofcurvatures ofthe surfaces are on the
For first surfece ifwe use the refraction formula as given below
Fli?2
Hi _ B. _ \
u
opposite sides ofthelenses andthese will have different signs. /=
R
Here wesubstitute w= inf,pj= 1,^2= M' H1
F=i?jandweget ...(5.97)
...(5.101)
Nowusingi?j = i?2 = R, weget R
f
2(nl)
...(5.102)
Figure5.230 shows thatfirst image/, isobtained afterrefraction Equation(5.102) gives only the magnitude of the lens. When from first surface ata distance v,,which canbetreated asobject we use this focal length in image formation by a lens then it is for the light raj« inside the lens falling on the secondsurface. substituted with appropriate sign. After refraction from second surface the final image is produced
at Fas all refractedrays converge at focal point oflens. Soifwe again use the refraction formula for refraction at the second
2. PianoConvex or PlanoConcave Lenses : These lenses
surface
forthis equation(5.99) willreduceto the formgivenbelow. 1^2
Hl U
R
have oneplane surface sowecantakeR, = inf and R^—R and
[Geometrical Optics
269
3. ConcavoConvex or ConvexoConcave Lenses: Such lenses have different radii ofcurvature and their centers ofcurvature
lie on the same side oflens so the signs ofthe two radii i?, and
As we have
= (111)
J
1_
Ri
Rt
willbesameandthe equation(5.99) willreduce tothe form given below
1
...(5.104)
/=
Above equations(5.101) to (5.104) students can keep on tips for quickreference forcalculation offocal lengthmagnitude of anythin lens while solvinga problemofimage formation.
5.15 Analysis of Image Formation by Thin Lenses
1
__1_ "/
...(5.108)
Equation(5.108) is called Lens Formula' used to find the
location of image produced bya thin lensfor a specific object forparaxial ra}^incident on the lens.
Aswehavealready discussed that for thinlenses magnitude of primary and secondary focal lengths of lenses are equal but their signs are different as these exist on two different sides of
the lens.Forthin lenses when a light rayincident on it then in For finding the exact location of imageproduced bya thin lens above lensformula wealways useits secondary focal length. If made upofamediumofrefi'active index p,wecanstartbyusing weuseincidentrayreference sign convention then always the refraction formula twice for refraction at the two surfeces ofthe focal length of a converging lens is positive and that of a lens. Figure5.231 showsa biconvex lens and a point objectis diverging lens is negative whereas in cartesian coordinate placed at a distance u from the optic center of the lens on its system sign convention it depends upon the direction from principal axis. Ifweanalyze the imageproduced duetorefraction which light ray incident on the lens.As alreadydiscussed that
atits first surface 5'j which isatlocation /j asshown infigure,
we will use the refraction formula for this. 1^2
B.
...(5.105)
R
U
Here we use m= «, v= v,^
= 1, ^2 = ^ and R= R^ (for first
surface) and we are not using signs as in general case lenses can be of different types and object can also be at different locations. This gives p
1
Vj
u
in this chapter we are using cartesian coordinate sign convention in solving different questions ofgeometrical optics. If students wish then theycan alsosolve each problem using the other convention also.
5.15.1 Lateral Magnification in Image Formation by a Thin Lens
Figure5.232 shows a small object (candle) placed on the principal axisof the lensat a distance u from the optic centerof
p 1 .
...(5.106)
the lens and it producesan image/ which is real and inverted.If
theheights ofthe object and image above theprincipal axisare h and ^'respectively then from the figure we can see that the angular sizes of both the objectand imageare same on the two
SiiRi)
sides of lens and it is equal to 9. 0
I
••
h
V Figure 5.231 Figure 5.232
Now for the light ray inside the lens which falls on the second
surface 52will again suffer refraction for which again weuse the
refraction formula as given in equation(5.105) with «=Vp v=v, Pj = p, i2= 1 and R  R^. This gives
l_ii V
Vj
triangles formed by objectheight and image height on principal axis we have h
lfi
...(5.107)
R.
u
/j' =
Adding equations(5.106) and (5.107), we get ^
1
/
Where
m
V
— \h = mh
Height of Image (A') Height of Ojbect (h)
v n
 Geometrical Optics
'270
Herem = (v/m) is called 'Lateral Magnification^ by a thin lens used to find the size of image and its orientation. In this relation of lateral magnification u and v are substituted with proper signs according to sign convention whichcan resultin positive and negative values of m. Positive value of m denotes that image is erected or on the same side of principal axis and negative value of m denotes that image is inverted or on the opposite side of the principal axis of the lens.
For small width object if image is produced by a thin lens (converging or diverging) then image width can be calculated by using the equation(5.109). But if objectsize is large then this relation cannot be used and in that case we need to calculate
the image of both edges ofthe object along principal axis and takethe difference as explainedin figure5.80 in ariticle5.9.5 for spherical mirrors. — 5.153 Variation Curves ofimage Distance vs Object Distance
5.15.2 Longitudinal Magnification by aThinLens
for a Thin Lens
Lateral magnification formula for thin lenses gives the image height above the principal axis of mirror and in this section we will discuss about the image width along the principal axis ofa thin lens. The relation in object and image width along the principalaxis ofmirror is called'LongitudinalMagnification'' as given below.
We have discussed the lens formula which relates the image distance fiom pole of mirror for a given object distance. The lens formula is given as
l_i
1
f~v~u ...(5.110)
v =
Longitudinal magnification ofimage is given as
+ 1
/ As focal length ofa given lens is constant and it can be positive
Width ofimage along Principal Axis m,
2f
Width of Object along Principal Axis
or negative depending upon the sign convention and type of lens used. Tfie above function given in equation(5.110) can be plotted as shown in figure5.234(a) and (b) for a converging and diverging lens with negative x direction on left and positive xdirection on right which we generally consider.
f
. V
1 Figure 5.233
Figure5.233 shows image formation of an object located at a distance x from the convex lens offocal lengthfwhich produces an image ofthis object at a distance^'which is real inverted and enlarged because object was placed between F and 2Fpoints. Here we can see that object edge A was close to C so corresponding image edge A' is also closer to C. Ifwe consider objectis of verysmall width dx and image produced is having a width dy then from lens formula we have
Real Image of / Real object /
Real Image of virtual object
II
+
u = j
u
0
u
Virtual image of real object
i_i =1 V
!
(a)
f
V
Here by coordinate sign convention we use ti= x,f= +f and v = +y
Diveiging Lens Image of Real
L 1
J_
f
X
y
/
Differentiating this expression we get . yf
0^ ——:rdx —ydy X
u+f
0
y
Virtual image of real object
y
Virtual object
/
From this relation we can get the 'Longitudinal Modification'' as
m, =
dx
2
..2 = m'
...(5.109)
(b)
Figure 5.234
Virtual image of virtual object
Geometrical Optics
271
5.15,4 Effectofmotion ofObject and Lenson Image
When object or lens is in motion the distance between object andlens changes which affects theposition and size ofimage. Tofind the imagevelocity and for analysis of image's motion we can differentiate the lens formula and find the rate at which
Solution
Let theradius offirst surface be/?, and refractive index oflens be p and parallel rays be incident on the lens. Applying refraction formula at first surface
distances between image and lensis changing. Ifweconsider a: andy as object and image distance from poleofmirroroffocal length/then by lens formula we have f
y
Differentiating the above relation withrespect to time, weget dx
\
dt
1
p1
Vi
00
/?.
...(5.112)
Refraction formula at second surface
X \
p
H
2p
V
V
20
...(5.113)
Addingequation(5.112) and (5.113),weget
dy dt
y
2
ilil + ^ V]
dx .
Where — is the relative velocity of object with respect to the
CO
R1 ^2p
V V]
/?,
20
dy
lens and — is the velocity of image with respectto lens. 11
^
dx
yR,
v„ = 
li1
2p
20
20
dy
' Vq ^vCusing — v^and —=v.) .2 *^0
20
mVr
1 . .1 2 1 1 1 2 =7(,na,r)+=~=
...(5.111)
v=40cm
Where m is the lateral magnification produced by the mirror. The expression of image speed as given in equation(5.111) is 10 20 valid only for the velocitycomponent of the image and object along the principal axis of the lens. Ifthe objectand mirror is in Thus all parallelraysincidenton the lenswill focus at a point motion along the direction normal to principal axis we can 40cm from the lens in the medium with refractive index 2. directly differentiate the height of object and image above # Illustrative Example 5.61
principal axis which are related as
h. = mhQ
A biconvex lens has focal length 50 cm and the radius of
Differentiating this with respectto time weget dhj
curvature of one surface is double that of other. Find the radii ofcurvature ifrefractive index of lens.material is 2.
dh. ~m
dt V.
m
dt •mv
Solution
OiV
dh:
Here we can use —7^= v.., and dt
dt

which are the
velocity components of image and object respectively in direction normal to theprincipal axis.
/?l/?2 f=
(pl)(/?,+/?2)
And we are given with/?= 2R.
# Illustrative Example 5.60 Focal length ofa thin lens in air, is 10 cm.
For biconvex lens focal length of lens is given as
/? = 20 cm
/=
Now medium on one side ofthe lens is
replaced by a medium of refractive index p = 2. The radius ofcurvature of surface of lens, in contact with the medium, is 20 cm. Find the point on /,ir= principal axis where parallel rays Figure5.235
incident on lens from air parallel to axis will converge.
50 =
2/?.
2/?.
(pl)(3/?,)
3(pl)
2^1 3(21)
R^ =75 cm /?2150cm
Geometrical Optics
# Illustrative Example 5.62
Solution
A convexlensoffocal length20 cm is placedat a distance5 cm
Due toreflection bya mirror, image ofobject isformed onitself when reflected ra>^ fells normally onthe mirrorandretrace the
from aglass plate =2J
^ ^
path ofincident rays. For this the image produced by thelens
placed at a distance 30 cmfrom lensonthe othersideofglass plate. Locate the final image produced bythis optical setup.
must be formed at the center of curvature of the mirror as shown in ray diagram5.237. A = 20 cm
Solution
A = 10 cm
Figure5.236 shows the optical setup described in question
/? = 20 cm
and the ray diagram for image formation. Scm
h
3cm
»+•
"
60cm
Figure 5.237 Image
For lens formula, we use 30cm 60cm
1cmH
/=20cm v=15cm
Figure 5.236
For lens formula to be used in refraction by lens, we use w =  30cm
__L i _zi 15 X ~ 20
/= + 20cm
^
i _ J__J_
i1 =1
X ^ 15
V u f J
60
'm
x = 60cm
1_
v"^30 ~ 20 20x30 v=——=60cm
Shift of image due to refraction by the glass slab is given as S=t
43
20
ii
# Illustrative Example 5.64 A convex lens is held 45 cm above the bottom of an empty
tank. The image of a point on the bottom of a tank is formed 36 cm abovethe lens. Now a liquid is poured into the tank to a
depth of40cm. It is found thatthedistance oftheimage ofthe same pointon the bottom ofthe tankis 48 cm above the lens. Find the refractive index of the liquid. Solution
S= 3 1— = 1 cm Thus position offinal image = 60+ 1= 61 cm.
When the tank in empty and point object O is placed at the bottom ofthe tank, then for lens formula we use M=  45 cm and v = + 36 cm
# Illustrative Example 5.63
A diverging lens of focal length 20 cm is placed coaxially 5 cm toward left ofa converging mirror offocal length 10 cm. Where would an object be placed toward left of the lens so that a real image is formed on object itself.
Using lens formula, we have
l_i _ J_ V « "/ Mv
/= « —V
45x36
1620
4536
81
= 20 cm
jGeome^ical Optics
,
;
273
When the liquid ispoured in thetankto a depth 40 cm, then image is produced at 48cm above the lens so again for lens formula we use v'=+48cm
If u' bethedistance oftheobject O' for image to be produced
V^.
at 48cm from lens then we use I/,
J___l_
1
J
~V /
48
1_
Plane mirror
«. = 15cm
20
Figure 5.238
Solving we get
The image /j acts as the object for theplane mirror andafter reflection oflight ra}^ from the plane mirror final imageproduced
H'=34.29cm
is
The distance between liquid surface and lens is 5 cm
In a planemirror, the image formed is at same distance
at which objectis kept fromit and sizeremain same. Sothe final image is produced at a distance 30  10 = 20cm as shown in
Distance of O' from the surface is 34.29  5 = 29.29 cm which is the apparent depth of object below water surface at which when it is placed, image is producedat the specifiedlocation.
the reflectedraysbytwicethe angle at which mirror isrotated.
The refractive index of liquid is given as
# Illustrative Example 5.66
Real depth
40
Apparent depth
29.29
figure5.238 atanangle 90® totheprincipal axis asmirror rotates
A point object is placed at a distance of 0.3m from a convex lens of focal length 20cm which is cut into two halves each of
= 1.37
whichis displaced by 0.5 mm as shown in figure5.239. Find the position of the image. If more than one image is formed,
# Illustrative Example 5.65
find their number and the distance between them.
An object of height 4 cm is kept to the left of and on the axis of a converging lens of focal length 10 cm at a distance of 15 cm from the lens.Aplanemirroris placedinclined at 45® tothe lens
L0.5 mm
axis, 10cmtotheright ofthelens.Findtheposition andsizeof the image formed by the lens and mirror combination. Trace the path of the rays forming the image. Figure 5.239 Solution
Solution
LetAB be the objectplaced at a distance of 15 crh fromthe lens We have studied thateverypartofa lens behaves like complete as shown in figure5.238. We shall calculate the position of lens and produces a separate image soin this case two images image formedbythis lens in absenceofplane mirror. areobtained. Theimage formation is shown in figure5.240.
For lens formula weuse «j= 15 cm, and /=+ 10 cm 1
j0.5mm
11321
^
vi ~ 10 15 ~ 30 ^ 30
or
v,=+30cm
So the image would be formed at 30 cm from the lens to the Figure 5.240
right ofit
The light rays from object passesthrough optical centres Magnification by lens is m, = — = — = 2 *' ' «i 15 ^
and OjUndeflected.The images of(9 are produced at/j and/2 due to upper and lower part of lenses. To find the location of
Sizeoftheimage = 2 x4 = 8cm
The image produced bylens is /, which is shown by figure5.238.
images we use refraction formula as
in
11=1 V
u
f
Geometrical Optics
i274
For refraction formula we use w=  30cm and/= + 20cm
I ^
I J_
V~
1_ J ~
l_
Am:
If d is the distance between the two images 7, and Ij then separation between these two images can be calculated from
Substituting the given values, we have
M+ V
Am
O1O2 30 + 60
Solving we get
(neglecting higher terms)
(vfY
triangle AOO, O2 and A7/j I2, inwhich by similarity, we use
0.05 + 0.05
(V/).
fAv
v=60cm
'
Av 1—
(V/)'
30 20 ~ 60
d
1
/'Av Am =
(25)^x18
(25)'xl8
» 0.5 mm.
(50025)2
# Illustrative Example 5.68
30
(7=3 mm
Determine the position of the image produced by an optical system consisting of a concave mirror with a focal length of 10 cm and a convergent lens with a focal length of 20 cm. The
# Illustrative Example 5.67
distance'from the mirror to the lens is 30 cm and from the lens
A thin converging lens of focal length/= 25.0cm forms the image of an object,on the screen, at a distance 5cm from the lens. The screen is then drawn closer by a distance 18cm. By what distance should the object be shifted so that its image on
to the object is 40 cm." Find the image produced after one refraction from lens and one reflection from mirror. Solution
the screen is sharp again?
Object is placed at distance 2/from the lens so its image is also formed at distance 2/on other side. For mirror this image acts as an object which is located at a distance of 10cm and we applymirror formula in this case for which we use m=+10cm
Solution
Since the image is formed on the screen, it is real 1
Now
1
and/=10cm
1
—+ ~ = "7 V
M
, /
1 =>
l_i
or
/ u =
U
V/
V
A V/
]_ J ...(5.114)
/(vAv) (vAv)/
1
V
J
r
1_
v"^+10"10
=>
In the second case, let the image is formed at v  Av. Let the corresponding position of object be m Am. Now. M+ Am =
^
1
 +
v=5cm
Raydiagram forimageformation is shown infigure5.241 below. 40 cm
...(5.115) 40 cm
The shift of the object m+ Am  m= Am
Subtracting equation(5.114)from equation(5.115),we get Am =
Am =
/(vAv)
fv
(vAv)/
vf
H5cm
/(vAv)(v/)>((vAv)/) {(vAv)/}(v/)
fAv Am =
(vAv/)(v/)
/^Av
Am =
(V/)'
!•
Av
(V/).
Figure 5.241
iGeometrical Optics
# Illustrative Example 5.69
which gives
U'=2{00)
= 2(^I)a/• A small angled prism(refractive index p and angle a) and a convex lens are arranged as shown in figure. Apoint object O Thus image position is3/ontheright side ofthe lens along the is placed as shown.
axis, and 2(p l)a/transverseto axis.
(a) Calculate the angle of deviation of the rays hitting the # Illustrative Example 5.70
prism at nearly normal incidence
(b) If the distance between object, prism and the lens are
shown in the figure, locate the position of the image both along and transverse to the axis.
Astrong source oflight when used with aconvex lens produces a number of images of the source owing to feeble internal reflections and retraction called flare spots as shown in
figure5.243. These extra images areF'j,
Solution
position of
. IfF^ is the
flare spot, then show that
The optical setup described inquestion isshown infigure5.242
(« + l)pl
below.
L
^4
F
/
/
^
7,f.
»
Figure 5.243
Figure 5.242
Solution
Figure change fromprintouts...
(a) The deviation produced bythe prism
Light converges at
after two refractions and one reflection
from the lens. So we use
5=(il)a
(b) The prism forms image ofthe objectat O'. 00' = bf={si\)af
J__i. h
"
fm
Where focal length ofequivalent independent lens isgiven do
The image O'becomes objectfor lens. 3/ Now using lens formula, we have u= —^ so we use
fe
V u~ f ^=2(piy
J_ I_ _ J_ V V "/
—~—+ F,
1 _ J_ V"3/
1
ir
2 2(pl)/
2pl
Fx (M1)/ For F^, there are three refractions and two reflections
v=3/ Also
f
V
i _ L A
OO' ~ u
F2 ~fx ^7^
IL IL 2
= 2 — +
1
1
/ Rn ^ f^R
Geometrical Optics^
5276
_1
By lens formula,
4
1 1=1 —, we u have _3_ '
V u
2 (lil)/
3(^l) + 2
3^11
(li1)/
(li1)/
1
1
V
0.5
f
v=0.6m
1
Web Reference at www.phvsicsgalaxv.com
(li1)/
Age Group  High.School Physics  Age 1719 Years
By using lens formula
Section  OPTICS
_1_' V w "/' 1
1
Topic  Geometrical Optics III Thin Lenses
h = + 15 cm,/= + 30cm
where
J_ we have
which gives
1
Module Number 1 to 14
1_
V +15 ~ +30
Practice Exercise 5.6
v=+ 10 cm
(1) A point sourceof light is kept at a distance of 15 cm from a converginglens, on its opticalaxis. The focal length of
The plot of rays is shown n figure.
the lens is 10 cm and its diameter is 3 cm. A screen is placed on
# Illustrative Example 5.71
the other side of the lens, perpendicular to the axis of the lens,
A small fish, 0.4 m belowthe surfeceofa lake, isviewedthrough a simpleconverging lensof focal length 3m. The lens is kept at 0.2 m above the water surface such that the fish lies on the
at a distance 20 cm fiom it. Find the area of the illuminated part of the screen ?
[(Tt/4) cm^l
optical axis of the lens. Find the imageof the fish seenby the observer. The refiactive index ofwater is 4/3.
(ii) A 5.0 diopterlensforms a virtual imagewhich is 4 times the objectplaced perpendicularlyon the principal axis of lens,
Solution
find the distance of object fi^om lens.
The apparent position ofthe object O fi"om the surface ofwater is
(Hi) A point object is placed at a distance of 25 cm fiom a convex lens of focal length 20 cm. If a glass slab of thickness t and refiactive index 1.5 is inserted between the lens and the
object, the image is formedat infinity. Find the thickness t ? [15 cm]
'
0.2m
(iv) A convex lens offocal length 20 cm and a concavelens offocal length 10cmare placed10cmapartwiththeirprinciple axis coinciding. A parallel beam of light of diameter 5 mm is incident on convex lens symmetrically. Prove that emerging beam will also be parallel & find its diameter. [2.5 mm]
Figure 5.244
The distance
PI = 0.3 + 0.2 = 0.5 m
For convex lens, we use
M=0.5m,/=+3m
(v)
A convexlens of focal length 20 cm is placed 10 cm in
fiont of a convex mirror of radius of curvature 15 cm. Where
should a point object be placed in front of the lens so that it produces image on itself? [100 cm]
iGeometrical Optics
__277J
(vi) A converging lensof focaMength 20 cm is separated 8 (xii) An object is kept at a distance of 16 cm from a thin lens cmfrom a diverging lens offocal length 30cm. Aparallel beam andtheimage formed isreal. If theobject is kept at a distance of light falls on converging lens and after passing through of 6cmfromthe samelens,the image formed is virtual. If the diverging lens focussed atpoint P. Find thelocation ofpoint P. sizeofthe images formed in above twocases are same,"findthe Repeat the calculation forthe casewhen theparallelbeam first focal length ofthe lens? falls on diverging lens. [ 11 cm]
[42.2 cm from convex lens]
(vii) Two symmetric double convex lensesA and B have same focal length but the radii of curvature differ so that R, = 0.9R„. A
D
Ifrefractive index ofA is 1.63 find the refractive index of.^. [1.7]
(vii!) An object is placed at 20 cm left ofthe convex lens of
(xiii) An object is placed midway between the lens and the mirror as shownin figure5.246. Themirror'sradius ofcurvature
is 20.0 cm and the lens has a focal length of  16.7 cm. Considering only the rays that leaves the object and travels first toward themirror, locate the final image formed bythis system. Is this imagereal or virtual? Is it upright or inverted? What is the overall magnification ?
focal length 10 cm. If a concave mirror of focal length 5 cm is placed at 30 cm to the right of the lens, find the magnification and the natureofthe final image. Drawthe ray diagramand locatethe position of the final image.
Y /
[At the object and of same size]
(ix) In the figure5.245 it is shown, the focal length/of the two thin convex lenses is the same. They are separated by a horizontal distance3/and their optical axes are displacedbya vertical separation 'if {d^is we can ignore such coefficients and most commonly we use the two term form ofCauchys Equationgiven as p=^+ —
(xi)
"t
...(5.147)
with pj = 1.70 andother diverging with pj = 151. Both lenses
With the above equation(5.147) wecanseethatlightwith lower
have same curvatureradius of their surfacesis equal to 10cm. The lenses were put close together and submerged into water. What is the focal length of this system in water.
travels slower. So if we compare violetand red light in same mediumthen we can seethat violetlight travelslowerthan red
wavelength has higher refractive index in a medium and hence
light in the medium. [33.3 cm]
Thisdifference in speed causes lightra)«ofdifferent wavelength (xii) A point object is located at a distance of 100 cm from a screen. A lens of focal length 23 cm mounted on a movable frictionless stand is kept between the source and the screen. The stand is attached to a spring ofnatural length 50 cm and spring constant 800 N/m as shown in figure5.277. Mass ofthe
to refract differently when these travel from one medium to another and that is the cause of the phenomenon called 'Dispersion ofLight\
stand with lens is2 kg. Howmuch impulseP shouldbe imparted
Figure5.278 showsa lightray of whitelight whichincidenton a transparent medium. As we know white light consist of all
to the stand so that a real image of the object is formed on the
colours sowecanconsider this lightrayas overlapping of light
screen after a fixed time gap. Also find this time gap also.
rays of all colours making the white light and for each colour
Geometrical Optics
1292
therefractive index ofthemedium will bedifferent and rayof colours but when this refracted beam emerges out oftheslab in each colour inwhite light will bend atdifferent angles and split airthen all the light rays inside the refracted beam will come out whenthe lightrayentersintothe medium as shown in figure.
at same angle ofemergence which isequal toangle ofincidence
so all'the rays become parallel after coming oiit and form a parallel white light beam with edges of violet and red colour only. Wecan alsoseethat dueto dispersion the emergent beam oflight is slightlythickerthan the incidentlight beam. glass
Figure 5.278
The situation shown in figure5.278 is an ideal situation which doesn't happen in general practice because of the dimensions of the light ray. Actually a light ray is a theoretical concept which does not have any dimensional thickness but in practice we consider a very thin light beam as a light ray. Look at the figure5.279 which is a magnified viewofthefigure5.278 in which a light beam (enlarged view of light ray) incident on the medium. In this beam we can consider several light rays incident on the medium boundary at the same angle ofincidence. Due to dispersion each light ray will split into its colours and you can see that except the light rays at the boundaries of the beam all the colours in between will merge into each other and produce white light again. Only the edges of light beam will have violet colour at the left edge in the medium and red colour at the right edge in the medium. This is the reason why at the time of refraction, colours are not seen in the refracted light,beam.
Figure 5.280
5.19.2 Dispersion of White Light through a Glass Prism When a beam of white light passes through a glass prism then due to two refractions of beam of white light the colours in
beamgets separatedfromeachother and at somedistance from theprismafterrefraction ifweplacea screen, coloured spectrum is obtained as shown in figure5.281(b).
Fkedfaeatn \ .
(a) y R f RV Ry
White Light Figure 5.279
5:19.1 Dispersion of White Light through a Glass Slab
Figure5.280 shows a light beam incident on a parallel sided glass slab at some angle of incidence. Due to the concept explained in previous section we can see that ihe edges of the refracted beam (slightly divergent) are having violet and red
(b) Figure 5.281
[Geometjical Optics 293
Figure5.281(a) shows how the coloured beams get separated beam. If the range ofwavelength is large then we take the after second refraction. We can see in this figure that at first difference ofrefractive indices corresponding to the wavelength
refraction at surface^5 ofthe prism, white light gets refracted limitsin the lightbeam. into the prism and the middle region between the violet and red,
edges will be white because ofmixing ofall the coloured light rays inthis region asexplained infigure5.280 but atthe second refraction when all these rays incident on the face AC of the prism then each colour at theincident region 'w«' ofbeam on
face will emerge out atthe same angle like all violet rays in this region 'mn' will emerge out atsame angle as aparallel beam and all red rays ofthis region emerge out as separate parallel
beam because both colours will have different emergent angles. This situation is unlike to the case discussed in figure5.280 when all coloured light rays emerge from the glass slab atsame angle and produce an emerging white light beam due to
Figure 5.282
overlapping of ail colours in same direction.
The concept ofdispersive power is used in taking corrective measures in image formation by refracting devices due to get separated after some distance and produce aspectrum on a dispersion. We'll discuss it later in the topic of 'Chromatic Figure5.281(b) shows practical situation inwhich allthe colours
screen. •
Aberratioh\
5.19.3 Dispersive Power of a Prism Material
5.19.4 Dispersion Analysis for aSmall Angled Prism
Dispersive power ofa substance is its ability to disperse light Figure5.283 shows a white light incident on a small angled passing through it.Dispersive Power ofa material is defined as
prism with prism angle^^. In general for awhite light we consider
relative deviation oflight beam from its mean path asafianction
its wavelength limits are bounded by violet and red rays and.
ofrefractive index ofthe material. For agiven material itis given mean wavelength is considered asyellow light. Iftherefractive
indices ofthe prism material for these colours are given as pp
as
...(5.148)
03 =
In equation(5.148) angle 0is the mean deviation oflight for all the wavelengths present initand af0 isthe change indeviation
p^and Pyrespectively and 5p 5^^ and 5yare the corresponding deviation angles when these colours pass through the prism then therelation among these are given below. ?>y=A{^y\)
due to variation ofrefractive index d\x which is corresponding
...(5.151)
to the wavelength change c(K in the beam. The dispersive power
for aprism material can be calculated by passing alight through
...(5.150)
by =A{^y\)
...(5.152)
a smallangled prismas shown in figure5.282. If weconsider a
light consist ofvery small wavelength range liol +dland itis passed through the small angled prism ofprism angle A as
shown, the mean deviation oflight for the wavelength Iis given as
e=^(n^i) r
/0 =Ad\x=
d\x LlI
Now from equation(5.148) thedispersive power ofmaterial is given as
dQ 03 =
Figure 5.283
d\}i p1
..(5.149)
Above equation(5.149) gives the dispersive power ofa prism material which ishaving the wavelength range dkinthe light
In above figure mean deviation of the incident light ray is considered corresponding to themean wavelength present in
the light so here mean deviation ofincident light is taken as 8y
^^Geom'^ricaLOptit^l 294

The 'Angular Dispersion' for the light passing through the For theabove case Which we call thesetup ofprism combination prism is defined as the angle bywhich light will totally disperse. for 'Deviation without Dispersion' or 'Achromatic Prism Combination'. Inabove case thetotal dispersion iszero hence
Hereangulardispersion is given as
the angular dispersion by the two prisms must be equal. IfX)
D—5j/^ 5j^ —A(pf/* 1)— =>
and D'are the angular dispersion by the first and second prism
D=
...(5.153)
thenthe condition of deviation without dispersion is given as D=D'
For the prism shown in figure average dispersive power isgiven
...(5.155)
as
D co=
5y
\Xy
Ap
P/j
Py 1
...(5.154)
Pavg 1
In above case thetotalmean deviation ofthewhite lightcan be given as
Note: In somecasesif meanrefractive indexcorresponding to
yellow light isnot given then students can use py
5™» = 5,5/
^ 5.19.6 Direct Vision Prism Combination
5.19.5 Achromatic Prism Combination
When two prisms of materials having different dispersive powers are placed in relative inverted positions with their prism angles are chosen in such a way that dispersion produced by one prism gets compensated by the other one then a ray of white light passing through this combination gets refracted
When two prisms ofdifferent materials are placed in contact with relative inverted positions and their refractive indices and
prism angles are chosen in such awaythat emerging light beam will not suffer any mean deviation but having dispersion in samedirection then such a prismcombination is called 'Direct Vision Prism Combination'.
and deviated but in emergent beam no colors are seen or no
overall dispersion of light beam take place then such a Figure5.285 shows combination oftwo prisms with inverted combination of prisms is called 'Achromatic Prism relative positions. The prism angles ofthese prisms are^ and ' Combination'. and having refractive indices for red and violet lights ij,and
}i^'respectively. Ifthese values ofprism angles and refractive Figure5.284 shows combination oftwo prisms placed incontact with opposite positions oftheir prism angles. The prism angles ofthese prisms are^ and^'.These prisms materials have their
indices are adjusted such that the mean deviation for yellow colour by first prism is exactly equal to that bysecond prism then the total deviation of the white light incident on this
refractive indices p^ and Pj^for first prism and p^'and p^'for combination willbezeroandthedispersed lightbeam will emerge the second prism corresponding to red and violet light out in the same direction of incidenceas shown in the figure. respectively. Ifthe values ofprism angles and refractive indices are adjusted such that the angular dispersion produced byfirst prism is equal to that produced by the second prism then the ray ofwhite light incident on the combination as shown in Yellow White Light figure will emerge out as while light only and will be free from colours. This happens because thenetdispersion produced by the first prism in the light is exactly nullified by the second
prism because itis placed inverted in respect ofthe first prism. Figure 5.285
'mean
White light
In above case ofdirect vision prism combination ifS^and fiy'afe the mean deviation oflightbythetwoprisms thenthe condition for this top happen is given as
A{v^y\)= A'{\x'\) Figure 5.284
...(5.157)
[Geomeffical'Optics^'';''f;; ,
1295:
In this case ifD and £)'are the angular dispersions produced by the two prisms.then the net angular dispersion ofthe emerging light beam from theprism combination isgiven as
...{5.158)
5.20 Optical Aberrations in Lenses and Mirrors
Optical aberrations are the defects in image formation byoptical devices dueto deviation in performance ofthe device because
of nonparaxial rays or polychromatic light used in optical systems. Aberrations occur because all light rays from one point object do not converge (or diverge) at a single point after undergoing reflection orrefraction through the optical device inuse. Inoptical systerhs the image produced gets blurred due toaberrations which causes unclear image fcamation bydifferent
(b) Figure 5.286
5.20.2 Methods to Reduce SphericalAberrations There are differentwaysby which correctivemeasurescan be
taken to reduce the spherical aberrations in image formation.
Some specific methods are given here for basic understanding, detailed analysis is not covered here.
corrective measures must be taken to avoid or minimize
(i) Use ofStops : Stops are the opaque planes having asmall aperture used to cutoffmarginal rays toincident on the optical
aberrations.
device as shown infigure5.287(a) and(b).
optical instruments. While designing optical instruments
Optical aberrations are classified in two broad categories in general. These are ^SphericalAberrations' and 'Chromatic Aberrations'. Spherical aberrations are due to the curvature of
optical devices which causes marginal rays not to meet at the point of image formation by paraxial rays; and chromatic aberrations are due tothe presence ofseveral wavelengths of light in theincident beam which causes different wavelength rays to refract differentlyfrom the optical device. Chromatic
aberrations areconsidered only in refracting optical devices as reflecting devices are free from chromatic aberrations.' 5.20.1 SphericalAberrations
(a)
Due to large aperture ofmirrors and lenses when marginal rays • incident on the outerpart ofthe mirrorsor lenses, thesedonot converge at the point where paraxial rays are meeting and
forming theimage. Figure5.286(a) and (b) shows theparallel light rays incident on a convex lens and a concave mirror. All
paraxial rays inthese cases will converge atfocal point butthe rays which are incident on outer edge' of the mirror or lens converge before focal pointand causes image to get blurred.
Due tothis reason a mirror ora lens having large aperture foils to produce sharp image ofan object or point image ofa point object.
•
• .• :
(b).
Figure 5.287
(ii) Using Parabolic Mirrors : For focussing parallel rays, spherical mirrors can be repla!ced by parabolic mirrors as
parabolic'mirrors focuses all parallel light rays foiling on itparallel (a)
toprincipal axis ata single focal point asshown'in figifre5.288'.
GeorrietricalOptics' 1296
i; i
i i
Figure 5.290
In above figure, the focal lengths/^and/^ are given as Figure 5.288
1
'1
__1_^
A
^2 ,
1
"sing this combination
for image formation, spherical aberrations are minimized.
I '
...(5.159)
fy
(iii) Using Convex Lenses at some separation : When two convex lenses areseparated bya distance equal tothedifference
in their focal lengths {d= \f^ /2I)
' 1
= (lifll)
/r
...(5.160)
Fromabove equations(5.159) and(5.160) weget
Using Crossed Lenses : Crossed lenses are specially designed lenses with their radii ofcurvature ofthe two surfaces chosen for a specific case of image formation in such a wayto
ir" A We can multiply the RHS numerator anddenominator bythe
minimizesphericalaberrations.
term (py1) where ^yis the refractive index for yellow (mean) colour in white light.
5.20.3 Chromatic Aberration in a Lens
As we have alreadydiscussed that chromatic aberrations are
due to presence of different colours (wavelengths) in the incident light beam which are refracted bythe optical device. Mirrors are free fromchromaticaberrationsas all coloursfollow
same laws ofreflection. Figure5.289 shows that a parallel beam
of white light falling on a concave mirror always produces a bright white spot atits focus whereas for a lens its focal length depends upon the refractive index oflens which isdifferent for different colours.
1
J__,
/
/r
_L
L
fv
/r
JJ
(Py1)
Rj
(11
Ir "fv _ i\^y
White light.
(Mrl)
(IVZM x(n 1) L1
1
fvfR
L
I Ad ^2 »J
^ 1
(liy1)
fy ...(5.161)
Ir fy ^fy Asfor mean colour wecan use/^yj; = fy
Equation(5.161) iscalled 'Longitudinal Chromatic Aberration' by a lens for white light. In general when a light consist of wavelengths ofsmall range from A. toX, + then corresponding
Bright white spot
variation in refractive index ofthe material oflens will be from p.
Figure 5.289
tod\i. Inthiscase thefocal length ofthelens for thelight can be
Below figure5.290 shows thesituation when a parallel beam of white light incident on the convex lens, due to dispersion it splits into colours and produces several coloured images because focal length of all colours of this lens given by lens
maker's formula will be different. As )iy>
bylensmaker's
formula we get/^
(5000)^
= 1.232
sin /2 = 1.232x0.8=0.9856
i2 = 80°16' Deviation of beam Q is given as
5^ =/2e =80°16'53°=27°16' (c) As the two beams have light wave of different wavelengths, these are non coherent light waves. When two or more non coherent light waves superpose each other on a point, the average resulting intensity is the sum ofindividual intensities of the component waves. Thus here the resulting intensity of the light at focus is 47+7 = 57.
(iv) A crown glass prism of angle 5° is to be combined with, a flint prism in such a way that the mean ray passesundeviated. Find (a) the angle ofthe flint glass prism needed and (b) the angular dispersion produced by the combination when white light goes through it. Refractive indices for red, yellow and violet light are 1.514,1.517 and 1.523 respectively for crown glass and 1.613,1.620 and 1.632 for flint glass. [(a) 4.169°, (b) 0.0348°]
(v) Fora crown and flint glassfor CandF'lines p^= 1.515 and p^= 1.523 andP(,=1.644, p^= 1.664 respectively. Calculate the angle of flint glass prism which may be combined with crown glass prism having refi^acting angle 20p such that the combination is achromatic for C and Frays.
Web Referenceat www.phvsicsgalaxv.com
(vi) The prism ofa spectrometer has a refracting angle 60° and is made ofglass whose refractive indices for red and violet are respectively 1.514 and 1.530. A while source is used and the instrument is set to give minimum deviation for red. Determine (a) angle ofincidence, (b) the angle ofthe emergence for violet light and (c) the angular width ofspectrum.
Age Group  High School Physics  Age 1719Years
[(a) 49°12', (b) 50°38', (c) TZb"]
Section  OPTICS
Topic  Dispersion of Light Module Number 1 to 12
(vii) An equiconvex lens of crown glass and an equiconvex lens of flint glass make an achromatic system. The radius of curvature of convex lens is 0.54 m. If the focal length of the combination for the mean colour is 1.54 m and the refractive
Practice Exercise 5,8
indices forthe crown glassare
= 1.53 and \Xy= 1.55, find the
dispersive power of the flint glass.
(i)
Two thin prisms are combined to form an achromatic
combination. For first prism ^ =4°, = 1.35, py=1.40, = 1.42. Forsecond prism p'^= 1.7, p'y.= 1.8 and p'^= 1.9find theprism angle ofsecond prism and the net mean derivation. [1.4°, 0.48°]
(ii) The index of refraction of heavy flint glass is 1.68 at 434 nm and 1.65 at 671 nm. Calculate the difference in the angle ofdeviation ofblue (434 nm) and red (671 nm) light incident at 65° on one side ofa heavyflint glass prism with apex angle 60°. [2.8°]
(Hi) The dispersive power of crown and flint glasses are 0.03 and 0.05 respectively. The refractive indices for yellow light for these glasses are 1.517 and 1.621 respectively. It is desired to form an achromatic combination of prisms ofcrown and flint glasses which can produce a deviation of 1° in the yellow ray. Find the refracting angles ofthe two prisms needed. [4.8°, 2.4°]
[0.055]
(viil) How would you use two planoconvex lenses of focal lengths 0.06m and 0.04 m to design an eyepiece fiee fiom chromatic aberration. What will be its focal length and magnifying power for normal vision ? Will it be a positive or negative eyepiece ? [0.048 m, 5.2, Negative]
(ix) An achromatic lensdouble is formed by placing in contact a convex lens offocal length 20 cm and a concave lens offocal length 30 cm. The dispersive power ofthe material of the convex lens is 0.18.
(a) Determine the dispersive power of the material of the concave lens.
^
(b) Calculate the focal length ofthe lensdoublet. [(a) 0.27, (b) 60 cm]
[^bmetrlcal Optics' 5.21 Optical Instruments
3or
Posterior chamber
Conjunctiva Sclcra
Optical Instruments are the devices vvhich utilize the
phenomenon ofreflection and refraction for image formation of various objects for their study indetail. All optical instruments aregenerally categorized intwo groups. One group ofdevices are those which produce real images of an object which is projected on ascreen oraphotographic plate and such images can be viewed simultaneously by many observers. Many different types ofProjectors and Spectroscopes belong tothis categorywhich we will not discuss in details as their construction
and working is not in scope ofthis book. The other group of devices are those which forms virtual image ofan object and only one observer can see the image. The virtual image formed by theinstrument istransformed by observer's eye into a real
Anterior chamber
Chomid
Retina
Tear?lm
N^treous
Cornea
Ciliary body, Optic nerve Canal of Schiemni Meibomian
glend Epithelium
Stroma
Endothclium
image on its retina. Some such optical instruments are also called as ' Optical Aids'" and such instruments are also used to
correct defects in human eye. Most commonly used optical instruments in this group are spectacles, microscopes and
Bowman's
Dcscemet's
membrane
membrane
telescopes which we study in detail. 5.21.1 TheHumanEye
In all types of optical instruments the observer's eye is an essential partasthe optical instruments we are going to study in coming sections, the analysis is based on observation of final image byobserver's eye so a good knowledge of human
eye is very important however in previous grades you might have studied it; Still for a quick reference we arediscussing the samehere again. Figure5.293(a) showsa sectionalviewofthe (b) human eye which isnearly spherical in shape having an average Figure 5.293 diameter of2.5cm for an adult and figure5.293{b) shows the front view ofa normal human eye. Therough outer protective When light rays enters into the eye through the cornea and and stiffskin ofeye ball is called 'Solera". Aithefront portion gets refracted from the eye lens and produces the image on a of eye, the sclera extends into a thin transparent membrane thin layer ofsensory cells inside the sclera. This thin lining is
which is called 'Cornea" in front ofthe eye lens. Justat theback called 'Retina" which acts asa screen for image formation. The
ofcornea isthe 'Iris", which isa textured pattern muscular ring.
surface ofretina ishemispherical inshape and it contains light
Iris can be of different colours which causes different color of
receptor cells called 'Rods" and 'Cones". These cells sense the
eye indifferent people. Atthe center ofiris there isan aperture imageproducedon retina and transmit it to human brain via the through which light incident on theeye lens. This aperture is 'Optic Nerve" shown in figure5.293(a). called 'Pupil" anditsdiameter isvariable andit gets broader or shrink toadapt tochanging light intensity falling on eye. Behind the iris there is a crystalline 'Eye Lens" which is biconvex and made up of fibrous jelly. The eye lens is soft at the edges and
For an object to be seen sharply, the image must be formed
exactly at retina. For different positions of object, eye adjusts
to different object distances bychangingthe focal length of its lensj the distance between lens and retina does not change. hard atthecenter. This eye lens isheld bycircular ringshaped The focal length of eye is adjusted by varying the radius of 'CiliaryMuscles" at theouter edge in front oftheeye ball. The curvature of the lens by the ciliary muscles. This process by' region between theeye lens andcornea contains a liquid called which ciliary muscles changes the curvature of eye lens to 'Aqueous Humor" and behindthe lens eyeis filled witha water obtain sharp image of different .object is called based jelly type liquid called ' VitreousHumor".
'Accommodation".
 ^Geornetricaf'Optics]
Human eye has limited capacity ofaccommodation which isin therange between ^Near Point' and 'Far Point'oftheeye. For a normal eye itsfar point is infinity. When eye focus.an object located at infinity then in this state ciliary muscles are fully relaxed. When eye focuses on an object placed at a short distance then the state at which ciliary muscles are fully
to eye itappears large and ifitisdisplaced away its size appear small dueto decrease in angular sizeof the object which it is
subtending on eye. Figure5.295(a) and (b) shows an object of size 'h' which islocated atdistances x, andxjffom an observer's
eye. The angular size ofthe object as seen by eye in the two situations is given as
•
h 0, = —
contracted and eye lens produces sharp image on retina, this distance is called near pointoftheeye. Thisminimum distance at which an eye can see objects distinctly and clearly without getting tired is called the 'Least Distance ofDistinct Vision'
and
1 Xi
.
^ h 0, = —
X2'
In
which is about 25cm for a normal eye. The range of
accommodation (near point tofar point) ofahuman eye gradually diminishes with ageandwith lifespan nearpoint andfarpoint
(a)
changes which results in defects of human eye called 'Presbyopia' and 'Myopia' which you have already covered in your previous grades. (b) 5.21.2 Camera
Figure 5.29^
Aphotographic camera isanoptical device which issimilar to Here angle 02 is less than 0, in the two positions ofthe same human eye in which image ofan object isproduced bya convex object and as we know the object which is located far away lens (camera lens) on photographic film instead of retina. fiom us appear smaller thats why here angle 02 is less than Figure5.294 shows the structural diagram ofa camera inwhich angle 0j. through the aperture light passes into the camera for short duration and produces a diminished real image on the photo sensitive film. The duration for which aperture opens and
5.21.4 Simple Microscope
captures the image decides the illumination ofthe image on N'Simple Microscope' or aG/as5'is aconverging film. lens of small focal length. When a magnifying glass, is held close to the object and.observer's eye is placed on the other side then in this situationthe size of virtual imageproducedis Shutter
larger then object and its distance isalso ferther firom the object as shown in figure5.296(b). In this figure the angular size of image is same as that ofobject which is 0 as shown.
i (a)
Aperture
r
h'
Figure S.294
1 ^
5.21.3 Angular Size of Objects and Images
.f ^
%]•
hwA:
•v = D
4
Whenthe observer seesan object directlyor byusing an optical instrument then the size of image appears to observer's eye is
analyzed byangular sizeofimage which also depends upon the distance where the image is located. Foran object located close
(b)
4
Georri'etricdOpticsj
•
magnified inverted image i4'5' at.a distance vas shown. This Thus magnifying power ofthe compound microscope in normal image A'B' is obtained atadistance from the eyepiece which conditions is given as is less than focal length of the eyepiece so that it produces a
virtual magnified image CD atadistance equal tonear point of the observer's eye which is kept close to the eyepiece on the
=>
other side for viewing the image.
^n=7.
14 —
...(5.176)
fs.
If in a compound microscope final image is produced at far
Eyepiece (X).
point ofthe eye then the magnification ofeyepiece is given by equation(5.173)as
Objective (/J
D
fE
Thus magnifying power ofthe compound microscope for final imageat far.point ofeyeis given as
...(5.177) Figure'5.297
Figure5.298 shows the situation when theimage produced by the objective A'B' isatthe focal point ofeyepiece due towhich thefinal image seen bytheobserver isatitsfarpoint (infinity).
The distance between two lenses, objective and eyepiece of
the'compound microscope is called Tube Length. We can'
Eyepiece (X)
X
5.21.8 'RibeLengthofa Compound Microscope
calculate tube lengthofa compound microscope in twd.cases
4\
when itproduces final image atthenear point and atfar point of
Objective (X,)
the observer's eye.
CaseT: When Image is produced at NearPoint
Thissituation isshown infigure5.174 in which thetube length can begiven as
ufo
Figure 5.298
, %
...(5.178)
lifo
Inequation(5.178) all values are substituted inmagnitude as
5.21.7 Magnifying PowerofCompound Microscope
signs are already considered with symbols.
Overall angular magnification ofthecompound microscope is
the product oftwo factors. First isthe lateral magnification of CaseTI: 'When Image is produced at Far Point the objective, which gives the linear size ofits image ^'5'and this situation isshown infigure5.175 in which thetube length second factor is theangular magnification oftheeyepiece which is given byequation(5.173) andequation(5.174) in the two cases ofimageformation at near point and far point ofthe eye.
w/o ufo
The lateral magnification of the objective is given as 0 =   = "
fo ufo
...(5.175)
Fora compound microscope innormal conditions when image
is produced at nearpoint then magnification of eyepiece is given by equation(5.173) D
can be given as
...(5.179)
In equation(5.179) allvalues aresubstituted in magnitude as signs are already considered with symbols. 5.21.9 Refracting^^tronomicalTelescppe
A telescopeisused;to view heavenly bodies. As heavenly bodies are very far away fiom earth,: theysubtend very small angular size .on the observer's, eye looking at these so these
^Geprnetrical Optics 305
bodies appear very small in size. A telescope produces a For the normal adjustment oftelescope for final image obtained magnified image ofsuch bodies which is having large angular at infinityas shown in figure5.299 the angular size ofimage can size so that eye can see more details ofthese bodies in enlarged be taken as p whichis given as' view ofimage. Atelescope which uses a lens as an objective is CdWoA
h
Refracting Telescope'.
...(5.181)
A simple astronomical telescope consists of two lenses From equations(5.180) and (5.181), the magnifying power of objective and eyepiece. The objective is a convex lens oflong telescope for normal adjustment is given as
focal length and itforms the real image ofdistant object infocal plane ofthe objective. Figure5.299 shows the optical system ofa simple refiacting telescope. The image AB formed'by
M
^
= 
a
objective acts as an object for eyepiece and based on distance ofeyepiece from thefocal plane ofobjective itproduces avirtual
and enlarged image ofAB either atnear point ofobserver's eye orat far point(infinity).
...(5.182)
' /£. In equation(5.182) a negative sign is inserted as the image produced by objective is inverted. For the situation when
telescope produces final image atnear point ofobserver's eye asshown infigure5.300 the image size p' isgiven as ...(5.183) , «£
From equation(5.180) and (5.183), themagnifying power of telescope for image at near point ofobserver's eye isgiven as Figure 5.299
^
Above figure5.299 shows the normal adjustment of lenses in
telescope inwhich the final image isproduced at infinity and observer's eye is in relaxed state which is the most common
adjustment for a telescope used in general and figure5.300
Mp
is obtained at anear point Dfrom eyepiece offocal lengthy^ so bylens formula
is given as %
...(5.185)
Ur.—
at near point ofthe eye.
H*—fa—\_
Eyepiece  r •
...(5.184)
Infigure5.300, the distance w^is such that image byeyepiece
shows the situation when final image isproduced by eyepiece ^Objective (/q)
a
Substituting the value of
from equation(5.i85) in
equation(5.184) we get fo D/e
fo f1 ,
Figure 5.300
5.21.10 Magnifying Power ofa Refracting Telescope
...(5.186)
In equation(5.186) a negative sign is inserted as the image
In all cases of telescopes always object is considered located produced by objective is inverted. far away fioih the device so we onlydiscuss in terms of the angularsizeof object and image. As shown in figure5.299 or 5.21.11 T\ibe Lengthofa RefractingTelescope figure5.300 the angular size of object can be taken as a and first image AB is produced in the focal plane of objective, if The distance between two. lenses, objective and eyepiece of the telescope is calledits' Tube Length'. We can calculate tube AB = h then we can use length ofa refi"acting telescope in two cases when it produces final image at the near pointand at far point of the observer's _h_ fo
...(5.T80)
eye.
Geometrical Optics 306
CaseI: When Image is produced at Near Point
5.21.13 Terrestrial Telescope
Thissituation isshown infigure5.300 inwhich thetube length
Thisisalso an 'Astronomical Telescope'' which produces erected
can be given as
images. This is similar to a normal refracting telescope but it uses anerecting lens between objective and eyepiece asshown j
=/• + —
in figure5.302. The disadvantage ofthis telescope isits large ...(5.187)
In equation(5.187) all values are substituted in magnitude as
tube length duetowhich it is rarely used. Objective
signs are already considered with symbols. CaseII; When Image is produced at Far Point Erecting Lens (/,)
Thissituation isshown infigure5.299 in which thetube length can be given as
Figure 5.302
5.21.12 Reflecting Telescope
In a reflecting telescope the objective isreplaced with a large concave mirror. This is an 'Astronomical Telescope' which is used for astronomical observations. To study the details of distant astronomicalbodieslarge apertureofobjective has many
advantages. Large aperture causes brighter image with fine details intheimage produced andmounting oflarge sized mirror ismuch easier compared tomounting ofa large lens. Figure5.301 shows theworking ofa reflecting telescope innormal adjustment
for obtaining the final image atinfinity. We can see that thefinal image in a reflecting telescope is formed in theregion where incoming rays are traveling in tube of telescope by placing a
Thetube length ofa Terrestrial Telescope is L=fQ + 4/,
Where/, is the focal length ofthe erecting lens. 5.21.14 Galilean Telescope
This isone among theoldest known telescopes made by Galileo Galilei in 1609.This uses a concave lens as eyepiece as shown
in figure5.303 which shows the normal adjustment of the telescope. The advantage ofthis telescope isshorter tube length and it gives erected image without using theerecting lens.
small mirror at an angle 45® to the axis of tube and placing eyepiece at the side of the tube.
H
fo
Figure 5.303
# Illustrative Example 5.87
Figure 5.301
A shortsighted man, the accommodation of whose eye is between 12cm and 60 cm wears spectacles through which he can see remote objects distinctly. Determine the minimum distance at which the man can read a book through his
Refracting telescope uses a thin lensas objective and lenses of spectacles. largeaperture cannot beproperly manufactured. If made then due tochromaticand spherical aberrations distorted imagesare Solution formed. Duetouseofparabolic mirrors asobjective inreflecting
telescope these arealmost free from chromatic and spherical aberrations.
As perthe given situation, a man can manage to see objects clearly ifplaced between 12cm and60cm (accommodation of
{Geometrical Optics 307
eye). If v is the distance between eye lens andretina then the 54
focal length of eye lens when an object is placed at 60cm distance is given by lens formula, used as
13.5
1 60
M=
...(5.189)
L
25
X (54/131) TT 2.5 =327.5
Ifheuses spectacles with focal length ofitslens f then hecan # Illustrative Example 5.89 see far objects clearly, that means for far objects the combination of eye lens and spectacles lens produces the Atelescope has an objective offocal length 50 cm and eyepiece
image at retina at distance v from eye lens then for the of focal length 5 cm. The distance of distinctvision is 25 cm.
combination ofthe lenses, we use lens formula as
i1 =L 1
The telescope is focussed for distinct visionon a scale 200 cm awayfromthe objective. Calculate ...(5.190)
^ ^ ///
From equation(5.189) and (5.190), weget
(i) The separation between the objective and eyepiece, (ii) Themagnification produced.
/=60 cm
Solution
For the near point ofthe eye at 12cm ifthe focal length ofthe The situation isshown infigure5.304 with raydiagram,
eye lens is^^' then bylensformula we use 1 12
...(5.191)
/;
(i) Ifthe separationbetween the two lenses bex then for lens formula forrefraction at objective lenswe use
WQ =200cm and /Q =+ 50cm
Ifthe minimum distance at which the person can read a book
clearly isplaced at a distance/) from eye then with spectacles
From lens formula, we have
lenses, we use lens formula as
]_
1
1
J—— = J
1_
^0
"o «o
To
...(5.192)
V ~D~ f' 60
J
1_
—+—
Fromequation(5.191) and(5.192), weget
/o
50
«0
41
/) = 15cm.
Illustrative Example 5.88
3
200 •
^0
200
200
200
Vo =+ 3
cm
The focal lengths ofthe objective and eyepiece ofamicroscope Thus a real image is formed at a distance of 200/3 from the are4 mmand25mmrespectively, andthelength ofthetube is objective. This image acts as object for the eyepiece. For 16 cm. If the final image is formed at infinity and the least
refraction through eyepiece, we use
distance of distinct vision is 25 cm, then calculate the magnifying power of the microscope.
^0" =_
X
3
v^ = 25 cmandj^= +5 cm
Solution
J
1
Magnificationof microscope is given as
I
200 "l ~ 5
25
X
M=^
£
"o
Here
v^, = 162.5= 13.5 cm
and
j^=+4mm =+0.4cm
Objective
*
200cni
Using the lens formula, we have
^0
^0
/o
Figure 5.304 X
Geometrical Optics' 308
Let Vo'be the new distance ofthe image formed bythe objective thenbyusing lensformula again, wehave
_i _L1
200 ^ ~ 5 25 ~ 25 1
X
_J_ J_ _ _i 1  L
3"^5~15
Vo'
6x400=25
VQ'=+15cm
6x=425 . .
Thus the image isshifted fiom the objective through adistance
425
15 cm6cm = 9cm. Sothe eyepiece shouldbe movedaway
x= ~r~ =70.80.cm 6
from the objective by 9cm to refocus the image atsame position. 200
(ii) Magnification of Objective
3x200
V.
3
25x6
Magnification of Eyepiece =
= 6
#Illustrative Example 5.91 In a compound microscope the objective and eyepiece have focal lengths of0.95 cm and 5 cm respectively, and are kept at a distance of 20cm.The final imageis formed at a distance of
Total magnification^
25cmfrom eyepiece. Calculate theposition oftheobject and
^ ^ 2
the total magnification. ' Solution
#Illustrative Example 5.90
Acompound microscope is used to enlarge an object kept ata
From the lens formula for eyepiece, we use
distance of 0.03 m from its objective which consists ofseveral
Ve =25cm andX"•'"5cm • e .
convex lenses in contact and has focal length 0.02m. Ifa lens
1 __i___i
offocal length 0.1 misremoved from theobjective, find out the distance by which theeyepiece ofthemicroscope must moved
^ " V, X " 25 5 " 25 =>
to refocus the image.
Li = A
Wp =(25/6)cm
For the objective, we use Solution
X=0.95cm and VQ=20(25/6) =95/6cm
Sol. Let Vq be the distance ofthe image formed by the objective Using lens formula, we have alone so by lens formula for objective, we use
Vo
"0
Vq
Wf)
/o
Vq
/o
Mq
/o
_1
1__^_1_
Vq
Wo
\
I
Vq
3 2
/o Wo
6
5
1
6100
0.96
95
_94
Vp=+6cm Wq
Theimage isformed at 6 cmbehind theobjective. If /q' bethe new focal length of the objective when a lensof focal length
95 95
«D='5Jcm
0.1 m(lOcm) is removed from it,then forcombination, we can use
/o'
_1_
1
_1_
/o
10
10
/;=+(5/2)cm
Total Magnification
M=^ Wo
1.^ L
(95/6) M=
(95/94)
1.^ 5
=94
{Geometrical Optics
309:
# Illustrative Example 5.92
If/be thefocal length ofthe correcting lensfor seeing distant objects then by lens formula for combination, we have
The focal lengths of the objective and the eyepiece of an astronomical telescope are 0.25 m and0.02m, respectively. The ...(5.194) V CO /^ / telescope is adjusted to view an object at a distance of 1.5m from theobjective, thefinal image being 0.25m from theeye of Subtracting equation(5.193) from equation(5.194), weget the observer. Calculate the tube length of the telescope and /=2m
the magnification produced by it.
Thus power ofthe lens required is Solution
1
1
P = — = — =Q.5 dioptre For objective, by lens formula, we have
J Vn
I_ 1.5
l__
.Web Reference at www.Dhvsicsgalaxy.com
+0.25
I Age Group High School Physics  Age 1719 Years
;r=="333 =>
Section  OFnCS
Topic  Optical Instruments Module Number  1 to 20
Vg=+0.3m
For eyepiece, by lens formula we have Practice Exercise 5.9
1
1
1
0.25
u,
+0.02
=>
1
I
1
Mg
0.25
0.02
(!)
Find (i) the size ofthe picture on the screen and (ii) the ratio of illumination of the slide and of the picture on the screen.
Mg=+0.01852m
The tube length of the telescope is given as
[2401]
1 = 0.3 + 0.01852 = 0.31852 m.
(ii) The focal length of the objective of a microscope is = 3mmandof the eyepiece^^ = 5 cm.An object is placed at a distanceof 3.1 mm fromthe objective. Find the magnification ofthemicroscope fora normaleye, ifthefinalimageisproduced at a distance 25 cm from the eye (or eyepiece). Also final the separation between the lenses of microscope.
Magnificationby objective is V,
A projector lens has a focal length 10 cm. It throws an
image ofa 2cm x 2cm slide on a screen 5 metre from the lens.
0.3
'"»=^=i?=" Magnification by eyepiece is 0.25
[180, 13.46 cm]
Total Magnification
= m, x Wj = 27.39
a Illustrative Example 5.93
A shortsighted person cannot see objects situated beyond 2m from him distinctly. What should be the power of the lens which he should use for seeing distant objects clearly? Solution
(iii) The optical powers of the objective and eyepiece of a microscope are equal to 100 D and 20 D respectively. The microscope magnification is equalto 50when imageisproduced at near point of eye. What will be magnification of the microscope be when the distance between the objective and eyepiece is increased by 2 cm ? [62]
(iv)
If^ be the accommodate focal length of the eyelenses for 2m, then by lens formula, we have 1 2
fe
...(5.193)
Where v is the distance between eye lens and retina.
The focal lengths of the objective and the eyepiece of
an astronomical telescopeare 0.25m and 0.025 m, respectively. The telescope is focussed on an object 5m from the objective, the final image being formed 0.25 m from the eyeofthe observer. Calculate the tube length of the telescope and its magnifying power.
[0.2859m, 11.6]
Geometrical Optics 1
(v) An astronomical telescope consisting of two convex lensesoffocal length 50 cm and 5 cm is focussed on the moon. What is the distance between the two lenses in this position ?
If the telescope is then turned towards an object 10 m away, how much would the eyepiece have to be moved to focus on the object without altering the accommodation of the eye? Calculate the magnification (angular) produced by the
(vii) An astronomical telescope in normal adjustment has a tubelength of93cmand magnification (angular) of30. If the eyepiece is to be drawn out by 3 cm to focus a near object, withthe final imageat infinity, findhowfar awayis the object and the magnification (angular) is this case. [27.9m, 31]
telescope in the two adjustments. [10]
Advance Illustrations Videos at www.phvsicsgalaxv.com
(vO The eyepiece and objective of a microscope, of focal lengths 0.3 m and 0.4 m respectively, are separatedbya distance of0.2 m. The eyepiece and the objective are to be interchanged such that the angular magnification of the instrumentremains same. What is the new separation between the lenses ? [0.2575 m]
AgeGroup  Advance Illustrations Section  OPTICS
Topic  Geometrical Optics Illustrations  46 Indepth IllustrationsVideos
Geometrical Optics
311
Discussion Question Q51 What is the function of the circular stop at the focal
Q516 How focal length of a spherical mirror changes when
plane of the objective of a telescope? Give reasons.
placed in different media.
Q52 Abeam ofwhite light passing through ahollow prism Q517 An air bubble inside water broadlybehaves as aconcave gives nospectrum. Isthis true orfalse? Give reasons. Q53 The magnifying power of a telescope in normal adjustment is greater than that when it is focussed for least
js
qj. f^igg .
Q518 Why does an aeroplane flying at a great altitude not cast a shadow on the earth ?
distance of distinct vision. Is this true or false? Give reasons.
Q54 Can youthink ofa specific opticalsetupwith a trihedral prism (no other device) in which a light ray passes undeviated through the prism. Think and draw the ray diagram. Q55 Can a real imagebe photographedbya camera ? Q56 Can you obtain image produced by a convex lens on a
Q519 The magnifying power of a telescope in normal adjustment is greater than that when it is focussed for least distance of distinct vision. Is this true or false?
Q520 When sunraj^ passthrough a smallholein thefoliage at the top of a high tree, they produce an elliptical spot of light on the ground. Explain why. When will the spot be a circle?
screen without using any other device.
Q521 Ifa mirrorreverses right and left, whydoesn't it reverse Q57 It is difficult to thread a needle with one eye closed.
up and down?
Why?
Q522 Is it possible to photographa virtual image ? Q58 If a single lens is used to forman image, it is better to use a lensof largediameter,in which the outerparts near the rim are blocked off. Explain.
Q523 Is it possible for a given lens to act as a converging lens in one medium, and as a diverging lens in another?
Q59 What is the bestposition of the eyeforviewing an object
Q524 A cameralens is markedJJ\.8. What is the meaningof
through a microscope?
this mark?
Q510 Can the optical length between two points ever be less than the geometrical path between these points?
Q525 Some motor cars have additional yellow headlights.
Q511 Can two lenses of the same material produce achromatismwhen placed in contact ? Explain.
Q526 Why are lenses often coated with thin films of
Q512 Whyis the objective of a telescope of large focal length and large aperture?
Q513 A iiver inside the sea observes a ship on the water
Why?
transparent material ?
Q527 If there are scratcheson the lens of a camera, they do not appear on a photograph taken with the camera. Explain. Do the scratches affect the photograph at all?
surface. Does he finds the ship taller or smaller then its actual height above the water surface. Give reason and draw ray diagram to support your logic.
Q528 The sun seems to rise before it actually rises and it seems to set long after it actually sets. Explain why.
Q514 A plane projectoris projectinga sharp still image on a
Q529 Doesthe focal length of a lens depend on the medium
screen, the image consists of objects of several colours. Can
in which the lens is immersed? Is it possible for a lens to act as
you comment on sharpness of all these coloured objects in image.
a converging lens in one medium and a diverging lens in another
Q515 Why does the moon, purely white during the day,have
Q530 Explain why the use of goggles enables an underwater swimmer to see clearly under the surface ofa lake.
a yellowish hue after sunset?
medium?
Geometrical Optics i
;312
ConceptualMCQs Single Option Correct 51 A beaker containing liquid is placed on the table underneatha microscope which can be movedalong a vertical scale. The microscopeis focussed, through the liquid onto a mark on the table when the reading on the scale is a. It is next focussed on the upper surface of liquid and the reading is b.
54 Two planemirrors oflengthI. are separated bydistance/, and a man
is standing at distance L from the connecting
lineofmirrors asshown infigure5.306. AmanAf, iswalking in a straightlineat distance 2Lparallel tomirrors at speed u,then man at Owillbeabletoseeimage ofM, for total time:
More liquid is added and the observations are repeated. The corresponding readings are c and d. The refractive index of
IM,
I
liquid is: dcb+a
d~b
(A)
(B)
dc~b + a
'd^b dcb+a
bd
(D)
(Q dcb+a
52 An insect ofnegligible mass is sitting on a block of mass M, tied with a spring of force constant k. The block performs simple harmonic motion with amplitude^ in front of a plane mirror placedas shown in figure5.305. The maximum speedof insect relative to its image will be:
21
Figure 5.306 3L
4L
(B)
(A) '///////A
9L
6L
P)
(Q —
u
55 A point source has been placed as shown in the figure5.307. What is the length on the screen that will receive reflected light from the mirror ?
Figure 5.305
H
(C)
^
n
'7777777777777: H
53 Inside a solid glass sphere ofradius R, a point source of light is embedded at a distance x(x < K) from centre of the
4
IH
Figure 5.307
sphere. The solid sphere is surrounded by air of refractive index 1.0. The maximum angle ofincidence for rays incident on
(A) 2H
the spherical glassair interlace directly from the point source
(Q H
(B) 3H (D) None of these
is:
(A) cos' —
(B) sin 1 A
(Q cos\/
P) sir' dj
R
56 A thin lens offocal lengthfand its aperture has a diameter d. It forms an image of intensity I. Now the central part of the aperture upto diameter {dH) is blocked by an opaque paper. The focal length and image intensity would change to
(A)/2,//2
'CB)///4
(Q 3flMI2
P)/3//4
jGeometrical Optics
313
57 A ray of light fallson a plane mirror. When the mirror is turned, about an axiswhich is at right angletothe planeofthe mirrorthrough 20°, theangle between the incident rayandnew reflected ray is 45°. The angle between the incident ray and original reflected ray was; (A) 65° (Q 25°or 65°
(B) 25° (D) 45°
510 Aninfinitely long rectangular'strip isplaced on principal axis of a concavemirror as shown in figure5.310. One end of the strip coincides with centre of curvature as shown. The
heightofrectangular stripis verysmallin comparison to focal length ofthemirror. Then theshape ofimage ofstripformed by concave mirror is similar to a:
58 In the figure shown5.308, lightis incidenton the interface
between media 1(refractive index p,)and 2 (refractive index Pj) at angle slightlygreaterthan the critical angle, and is totally reflected.The light is then also totallyreflected at the interface
between media 1and 3(refractive index P3), after which ittravels in a direction opposite to its initial direction. The media must have a refractive indices such that:
Figure 5.310
(A) Rectangle (Q Triangle
(B) Trapezium (D) Square
media 1
511 A concave spherical surface of radius of curvature 10cm separates two mediums X and Yof refractive indices 4/3 and 3/2 respectively. Centre ofcurvature ofthe surface lies in the medium X. An object is placed in medium X: (A) Image is alwa>^real (B) Image is real ifthe object distance is greater than 90 cm (Q Image is always virtual (D) Image is virtual only if the object distance is less than
media 3
90 cm
Figure 5.308
512 In the figure.shown5.311 a slab of refractive index —is
(A) p,pf
59 Figure5.309 shows a spherical cavityin a solid glassblock. The cavity is filled with a liquid and from outside an observer sees the distance AB which is the diameter of the cavity and it
moved at speed 1m/s towards a stationary observer. A point 'P' is observed by the observer with the help of paraxial rays through the slab. Both 'O' and observerlie in air. The velocity with which the image will move is:
appear as infinitely large to the observer. Ifrefractive index of
liquid is p. and that of glass is p,, then — is : ^2
Glass
Figure 5.311
(A) 2 m/s towards left
(B) —m/s towards left
(Q 3 m/s towards left
(D) Zero
513 Light passes from air into flint glass with index of refraction n. The angle of incidence must the light have for the component of its velocity perpendicular to the interface to Figure 5.309
(A) 2 (Q 4
(B) 1/2 (D) None of these
remain same in both mediums is:
(A) sin"'« (Q COS"' n
(B) sin"'(l/«) (D) tan"' n
Geomefricaj^ pities j
1^1:
514 Inthefigure5.312M, and are two fixed mirrors shown. Ifthe object' O" located between thetwo mirrors moves towards theplane mirror, then the image I which is formed after two successive reflections first from Mj & then from
517 Apoint object ismoving along principal axis ofaconcave
respectivelywill move:
the number oftimes at which the distance between object and its image is 40 cm are.
mirrorwithuniform velocity towards pole. Initially theobject is at infinite distance from pole on right side of the mirror as
shown inthe figure5.315. Before theobject collides with mirror,
object 0 z
6M,
Figure 5.312 Figure 5.315
(A) Always towards right (B) Always towards left (Q Depends on position of O P) Cannot be determined
p) Two times P) Data insufficient
(A) Onetime (Q Three times
515 A person AB of height 170cm is standing in fiont of a plane mirror. His eyes are at height 164 cm. Atwhat distance
518 In the figure shown5.316, the maximum number of
firom P should a hole be made in the mirror so that he cannot
reflections light rays will undergo are: M,
see the top ofhis head :
B
Figure 5.316
P
Figure 5.313
(B) 161an
(A) 167cm (Q 163cm
(B)3 P)1
(A) 2 (Q 4
P) None of these 519 A convex lens is cut into two parts in different ways that
516 In the figure shown5.314, blocks P and 0 arein contact but do not stick to each other. The lower face ofP behaves as
are arrangedin four manners, as shown. Which arrangement will givemaximum opticalpower?
a plane mirror. Thesprings are in theirnatural lengths. The system is released from rest.
(A) I
Ua
(Q
Figure 5.314
Then the distance between Q and its image when Q is at the lowest point first time will be: 2mg
(A)
K
^mg (Q
K
4OTg
(B) P)0
K
520 A parallelbeam of light passes parallelto the principal axis and falls on one face ofa thin convex lens offocal length/ and after two internal reflections fi"om the second face forms a
real image. The distance of image fiom lens if the refiactive index ofmaterial oflens is 1.5 :
(A)//7 {€) If
.
P)//2 • p) None of these
jGeometrical Op^cs
521 An object and aplane mirror are shown in figure5.317.
524 The position ofareal point object and its point image are
Mirror is moved with velocity Vas shown. The velocity of asshown in lhefigure5.320./15 istheprincipal axis. Thiscan be achieved by using :
image is:
Or
• Object (fixed)
T a
A
[ 1
^
a
i
Mirror
Figure 5.320
Figure 5.317
(A) 2Ksin0
(B) IV
(Q 2Fcose
P) None of these
522 A prism ofangle ^4 and refractive index 2 is surrounded
bymedium ofrefractive index >/3 .Aray is incident on sidePg at an angle of incidence / (0 < / < 90®) as shown in the figure5.318. The refracted ray is then incident on side PR of prism.The minimum angle^4 ofprism forwhichrayincidenton
sidePQ does not emergeout of prism from sidePR (for any value of/) is:
(A) Convexmirror (Q Planemirror only
(B) Concave mirror
p) Convexmirror only
525 The image of the moon is produced bya convex lensof focal length/ Thearea ofimage is directly proportional to: (A)/ (B)f (Q 1// P) yf ' 526 A converging lens of focal length 20 cm and diameter 5 cmis cutalongthe VmeAB. The part ofthe lensshown shaded inthe diagram isnowused to form an image ofapoint/*placed 30 cmawayfrom it on the lineA7, whichis perpendicular tothe plane ofthe lens. The image ofP will be formed:
Normal
T 2 cm
2^
a Incident ray 5 cm
X
•30 cm
Figure 5.318
(A) 30® (C) 60®
(B) 45® (P) \2QP
523 Two plane mirrors are joined together as shown in figure5.319. Two point objects O, and are placed symmetrically such that AO^ = AO2. The image of the two objects is common if:
Figure 5.321
(A) 0.5 cm above AT (C) on AT
P) 1 cm belowAT P) 1.5 cm belowAT'
527 A man stands on a glass slab of height h and inside an elevator accelerated upwards with 'a'. If is refractive index of glass then the bottom ofthe slab appears to have shifted with respect to the man by a distance ; (A) less than P)' greater than (Q equal to h/\i P) can't be said
528 When anobject isata distance m, and
from theoptical
centre ofa lens, a real and virtual image are formed respectively,
with the same magnification. The focal length oflens is: (A) (u. + M,) Figure 5.319
(A) 0 = 60= (Q 0 = 30=
(B) 0 = 90= (P) 0=45=
(Q
«+«2
P) «,+ y «i —U
P)
Geometrical
5316
529 A liquidof refractive index 1.33 is placed between two 532 Aplane mirror having amass mistied tothe free end ofa identical planoconvex lenses, with refractive index 1.50. Two massless spring of spring constant k. The other end of the
possible arrangement P and Qareshown infigure5.322. The spring is attached to a wall. Thespring with the mirror held vertically tothefloor onwhich itcanslide smoothly. When the system is: spring isatitsnatural length, themirror isfound tobemoving at a speed of v cm/s. The separation between the images of a man standing before themirror, when themirror isin itsextreme positions:
Wall
Figure 5.324 '
Figure 5.322
(A) Divergentin P, convergent in Q (B) Convergent in P, divergent in 2 (Q Convergent in both QD) Divergent in both
\m
V
(A) Vjy \m
(Q 2vJj
m
m
(D)4vj
530 Twoparticles^&Pofmass Wj and^2 respectively start moving from O with speeds v, and V2. A moves towards the 533 Anobserver cansee, through a pinhole, thetop endofa planemirrorand B moves parallelto mirrorhorizontally. The' thin rod of height h, placed as shown in the figure5.325. The mirror is inyz plane.The absolutespeed ofimageof centre of mass of the system (image of ^4 + image ofB) is:
beakerheight is 2h and its radiusis h. Whenthe beakeris filled with a liquid up to a height 2h, he can see the lowerend of the rod. Then the refractive index ofthe liquid is:
Figure 5.325
Figure 5.323
(A) 5/2
(B)
(Q JJ/T.
(D) 3/2
WiVi
(A) Zero
(B)
m2
534 A ray oflight isincidenton the leftvertical face ofa glass cube ofrefractive index
(Q
2^2
(D) None of these
m,
531 A mango tree is at the bank of a river and one of the
as shown in figure5.326. The plane
ofincident is the plane of the page, and the cubeis surrounded
by liquid of refractive index p,. What is the largest angle of incidence 0j for which total internal reflection occurs atthetop surface?
branch oftree extends over the river. A tortoise lives in'river. A
mango falls just above the tortoise. The acceleration of the mango falling from tree appearing to the tortoise is (Refractive index ofwater is 4/3 and the tortoise is stationary): (A) g
(B) f
4g (Q —
(D) None of these Figure 5.326
. Geometrical Optics
(A) sin 1 =
317
P) 45=
(Q 60°
p) 60°sin \lj
(B) sin 1 = /
(Q sinl
(A) 30°
(D) sin 1 =,
\2.
^1 +1 >^2
535 Twoplane mirrors are inclined to each other at angle 0. A ray oflight is reflected first at one mirror and then at the other. Find the total deviation ofthe ray: (A) 360°2e (B) 36O'' + 20 (Q 180°28 P) 18O'' + 20 536 Two plane mirrors are inclined to each other such that a ray oflight incident on the first mirror and parallel to the second is reflected fi^om the second mirror parallel to the first mirror. Determine the angle between the two mirrors: (A) 60° (B) 30° (Q 90° (P) 180°
539 The mirror of length L moves horizontally as shown in thefigure5.329 with a velocityv.The mirror is illuminatedbya point source of light 'P' placed on the ground. The rate at which the length ofthe light spot on the ground increases is : r*—L—H
bvWWWN
V,
Wall
Figure 5.329
p) Zero P)3v
(A) V (Q 2v
540 Monochromatic light rays parallel toxaxis strike a convex 537 A slab of high quality flat glass, with parallel faces, is placed in the path ofa parallel light beam before it is focussed to a spot by a lens. The glass is rotated slightly back and forth from the dotted centre about an axis coming out of the page, as shown in the diagram. According to ray optics the effect on the focussed spot is:
lenses ofrefractive index 0.5. Ifthe lens oscillates such that
AB tilts upto a small angle 0 (in radian) on either side ofyaxis, then find the distance between extreme positions ofoscillating image:
Figure 5.330
Rotating glass Figure 5.327
(A) There is no movement ofthe spot (B) The spot moves towards then away from the lens (C) The spot moves up and down parallel to the lens P) The spot moves along a line making an angle a (neither zero nor 90°) with axis oflens
538 Aparallel glass slab ofrefractive index >/3 isplaced in contact with an equilateral prism ofrefractive index Jl. Aray is incident on left surface of slab as shown. The slab and prism combination is surrounded by ain The magnitude ofminimum possible deviation ofthis ray by slabprism combination is:
(A) /secO (Q /(secO1)
P) / sec^O P) The image will not move
541 The focal length ofa concave mirror is/and the distance from the object to the principal focus is x. Then the ratio ofthe size ofthe image to the size ofthe object is: (A)
if + x) f
/ P) 
X
(D) ^ 542 A light ray gets reflectedfrom a pair ofmutuallyLmirrors, not necessarily along axes. The intersection point ofmirrors is at origin. The incident light ray is alongy = x+2. Ifthe light ray strikes both mirrors in succession, then it may get reflected finally along the line: (A)y2x2 p)y = x + 2 (Q y = x2 p)y = x4
Figure 5.328
Geometrical Optics
1318
543 A ray is incidenton the first prism at an angle ofincidence 53® as shown in the figure5.331. The angle between side CA and B'A' for the net deviation by both the prisms to be double of the deviation produced by the first prism, will be: A
i=53°
Figure 5.331
, 2
, 2
(A) sin' 2 +53'
(B) sin' 2+37'
(Q cos'2+53®
(D) 2sin"' j
♦
+
*
*
+
IGeometrical Optics
NumericalMCQs Single Options Correct 51 An equilateral prism deviates a raythrough 45® for two angles of incidencediffering by20®. The refractive index of the prism is:
(A) 1.567 (Q U
(B) 1.467 (p) 1.65
52 Two plane mirrors I, and
55 Two plane mirrors AB andAC are inclined at an angle 0= 20°. Arayoflight starting from point Pis incident at point 0 on the mirror AB, thenatR onmirror ACand again onSon AB. Finally the ray STgoes parallel to mirror^C. The angle which theraymakes with the normal atpoint Qonmirror AB is:
areparallel toeach other and
3m part. A person standings m fromthe right mirror looks into this mirror and sees a series of images. The distance between the first and second image is 4 m. Then the value ofx is:
y2^/777Z777777Z^7777^^^ A
PR
Figure 5.334
(A) 20® (Q 40®
vv
P) 30° .
\*—x
P)60®
56 A person's eye level is 1.5m. He stands in front of a 0.3m Figure 5.332
(A) 2m (Q Im
longplanemirrorwhich is0.8m above theground. Thelength ofthe image he sees ofhimselfis :
(B) 1.5m (D) 2.5m
53 An elevator at rest which is at 10''^ floor of a building is
(A) 1.5m (Q 0.8m
p) 1.0m P) 0.6m
having a plane mirror fixed to its floor. A particle is projected
57 A plane mirror of length 8 cm is present near a wall in situation as shown in figure5.335. Then the length of spot
with a speed V2 m/s and at45® with the horizontal asshown
formed on the wall is:
in the figure5.333. At the very instant ofprojection, the cable of the elevator breaks and the elevator starts falling freely. What will bethe separation between the particleand is image
,
wall
1 5 = source oflight d
i
0.5s after the instant of projection ?
V77777777/. H—8cm—H
Figure 5.335
u = 'J2 m/s
(A) 8 cm (Q 16cm
P) 4 cm P) None of these
58 Anobject 'O' is keptinair in front ofa thinplanoconvex lens ofradius ofcurvature 10 cm. It's refractive index is 3/2 and Mirror
Figure 5.333
the medium towards right of plane surfeceis water ofrefractive
index 4/3. What should bethe distance 'x' ofthe object sothat (A) 0.5m (Q 2m
(B) Im p) 1.5m
the rays become parallel finally. M
JT
»•
54 Aplane mirror is moving with velocity 4f+4/ +8^. A point object in front of the mirror moves with a velocity
1
'ii+Aj + Sk . Here k is along the normal to the plane mirror
«,3/2A
and facing towards the object. The velocity of the image is :
rtfz
^—
Figure 5.336
(A) 3^'4j +5A
(B) 3f+4y+llit
(Q ~4i+5j + Uk
p) 7i+9j+3k
(A) 5 cm (Q 20cm
P) 10 cm (D) None of these
Geornetrical Optics 
59 A diverging lens of focal length 10cm isplaced10cm in front of a plane mirror as shown in the
513 The curvature radii of a concavoconvex glass lens are, 20 cm and 60 cm. The convex surface of the lens is silvered. With the lens horizontal, the concave surface is filled with
figure5.337. Lightfrom a veryfar away source falls on the lens. The
water; The focal lengthof the effective mirror is (p of glass= 1.5, pofwater = 4/3):
I
final image is at a distance:
(A) 20cmbehindthe mirror
(A) 90/13 cm (Q 20/3 cm
(B) 80/13 cm p) 45/8 cm
Figure 5.337
7.5 cm in front ofthe mirror
514 Aplanoconvex lens, when silvered at itsplane surface isequivalent toa concave,mirror offocal length 28cm. When
(C) 7.5 cm behind the mirror (D) 2.5 cm in front of the mirror
its curved surface is silvered and the plane surface not silvered, 510 A fish in near the centre of a spherical fish bowl filled
it is equivalent to a concave mirror offocal length 10cm,then
with water of refractive index 4/3. A child stands in air at a
the refractive index ofthe material of the lens is:
distance IR {R is the radius of curvature of the sphere) from the centre ofthe bowl. At what distance from the centre would
(A) 9/14 (Q 17/9
the child's nose appear to the fish situated at the centre : (B) 2/? (A) AR (Q 3^ (D)4fi
515 A prismhas a refractive index
511 An objectis placedat a distanceof 15cm from a convex lens of focal length 10 cm. On the other side of the lens, a convex mirror is placedat its focus such that the imageformed bythe combination coincides with the object itself. The focal length of the convex mirror is : •15cm
10cm
H jp'
(B) 14/9 P) None
andrefracting angle
90°. Find the minimum deviation produced by the prism. (A) 40° P) 45° (C) 30° P)49°
516 A certainprism is found to produce a minimum deviation of 38°. It produces a deviation of 44° when the angle of incidence is either 42° or 62°. What will be the angle ofincidence when it undergoes minimum deviation ? P) 49° (A) 45°
(Q 40°
P) 55°
517 A light ray is incidenton a prism of angleA = 60° and
refractive index p= Jl •The angle ofincidence atwhich the Figure 5.338
(A) 20 cm (C) 15cm
(B) 10cm (D) 30 cm
512 Two planoconvexlenses each of focal length 10 cm & refractive index3/2 are placedas shown5.339. Water (p = 4/3) is filled in the space between the two lenses. The whole arrangement is in air. The optical power of the system in diopters is:
emergent ray grazes the surface is given by: (A) sin
(Q
sm
1
V31
I
2
P) sin
1
P) sin 1
\S 2_ 41
518 A transparent cylinderhas its right halfpolished so as to act as a mirror.Aparaxial light ray is incidentfrom left,that is parallel to principalaxis, exitsparallel to the incidentray as shown. The refractive index n ofthe material ofthe cylinder is :
Figure 5.339
(A) 6.67 (Q 333
(B) 6.67 (D) 20
Figure 5.340
(A) 1.2 (Q 1.8
P) 1.5 P)2.0
iGeometricai' Optics
321
519 Acomposite slab consisting ofdifferent media isplaced
30°.The refractiveindex ofthe liquid is :
in front ofa concave mirror ofradius ofcurvature 150cm. The
whole arrangement isplaced in water. Anobject O isplaced at
Point source
a distance 20cm from the slab. The refractive indices ofdifferent
mediaare givenin the diagramshownin figure5.341. Find the position of the final image formed by the systemf
M=4/3
H=4/3 \
M=1.5
f15;
ti'qu id
0
, .20 cm.
Figure 5.343 45 cm
24 cm
54 cm 10 cm
Figure 5.341
2_ S
(A) ^
1
(A) To the left of Object (B) On the Object
(Q iz
(C) To the right ofObject
523 In the figure5.344 ABC is the cross section of a right angled prism and BCDE is the cross section of a glass slab. The value of 0 so that light incident normally on the face AB does not cross the faceBC is (given sin"' (3/5) = 37°):
(Q Data insufficient to calculate the image position
520 A luminous point object is moving along the principal axis ofa concave mirror offocal length 12 cm towards it. When its distance from the mirror is 20 cm its velocity is 4 cm/s. The velocity ofthe image in cm/s at that instant is : (A) 6, towards the mirror (B) 6, away from the mirror (Q 9, away from the mirror (D) 9, towards the mirror
B
E
IN •o
fn
II
II
^ 5^.
521 Twoblockseach of mass m lie on a smooth table. They are attached to two other masses as shown in the figure5.342. The pulleys and strings are light. An object O is kept at rest on the table. The sides AB & CD of the two blocks are made
C
D
Figure 5.344
(A) 0 p^.
IGeometrical Optics
329
529 An object O is keptinfrontofa converging lensoffocal
(0 The average speed of the image as the object moves with
length 30 cm behind which there is a plane mirror placed at a distance 15 cm from the lens. Then which of the following
uniform speed from distance
3F
F
to — is greater than the
statements is/are correct.
average speed of the image as the object moves with same 30 cm
F
F
2
4
speed from distance — to —
(iQ The minimum distance between a real object and its real image in case of a converging lens is 4F where F is its focal
•15 cm
15 cm
Figure 5.369
(A) The final image is formed at 60 cm from the lens towards right ofit O) The final image is at 60 cm from lens towards left ofit (C) The final image is real (D) The final image is virtual
530 Choose the correctalternative corresponding to the object distance 'w', image distance 'v' and the focal length 'P of a converging lens from the following.
length. (A) both are correct (B) both are incorrect (Q (i) is correct, (ii) is incorrect (D) (i) is incorrect, (ii) is correct
531 An object and a screen are fixed at a distance d apart. When a lens offocal length/is moved between the object and the screen, sharp images ofthe object are formed on the screen for two positions of the lens. The magnifications produced at
these two positions areM, andM^. {A)d>2f
Af
(Q
(jy)\M,\\Mji=\
Geometrical Optics :
?330
Unsolved NumericalProblemsfor Preparation ofNSEP, INPhO & IPhO For detailedpreparation ofINPhO andIPhO students can refer advance study material on ww>v. physicsgalaxy.com 51 The left end ofa long glass rod ofindex 1.6350 is grounded and polished to a convex spherical surface ofradius 2.50 cm. A small objectis located in the air and on the axis 9.0 cm from the vertex. Find the lateral magnification.
ofthree thin lenseswith a common optical axis. The focal lengths
ofthelenses areequal to/j = 10cm (converging) and^ = 20cm (divering)and^ = 9cm(converging) respectively. Thedistance between the first and the second lens is 15 cm and between the
Ans. [ 0.0777] '
52 Focal length ofa convex lens in air is 10 cm. Find its focal
lengthin water. Given that
59 A parallel beamoflight is incidenton a system consisting
=3/2 and\x^ = 4/3.
Ans. [40 cm]
53 Find the distance ofan object from a convex lens ifimage is two times magnified. Focal length ofthe lens is 10 cm. Ans. [5 cm, 15 cm from lens] '
second and the third is 5 cm. Find the position of the point at which the beam converges when it leaves the system of lenses. Ans. [Infinity]
510 A ray of light is incident on the left vertical face ofglass cube ofrefractive index as shown in figure5.370. The plane of incidenceis the plane of the page, and the cube is surrounded
by liquid of refractive index pj. What is the largest angle of incidence 0] for which total internal reflection occurs at the top surface ?
54 A pole 4m high driven into the bottom of a lake is Im above the water. Determine the length of the shadow of the pole on the bottom ofthe lake if the sun rays make an angle of 45° with the water surface. The refractive index ofwater is 4/3. Ans. [2.88 m]
55 An object is placed 12cm to the left ofa diverging lens of focal length 6.0 cm. A converging lens with a focal length of 12.0 cm is placed at a distance dto the right of the diverging lens. Find the distance d such that the final image is produced at infinity.
Figure 5.370
Ans. [ sin"
{i
111
Ans. [8 cm]
56 A solid glass sphere with radius R and an index ofrefraction 1.5 is silvered over one hemisphere. A small object is located on the axis of the sphere at a distance 2R to the left of the vertex ofthe unsilvered hemisphere. Find the position offinal image after all refractions and reflections have taken place.
511 One face ofa prism with prism angle 30° is coated with silver to make it reflecting. A ray incident on another face at an angle of 45° is refracted and reflected from the silver coated face and retraces its path. What is the refractive index of the prism ? Ans. [ V2 ]
Ans. [On the pole of mirror]
57 A glass sphere with 10 cm radius has a 5 cm radius spherical hole at its centre. A narrow beam of parallel light is directed into the sphere. Find the location of final image produced ? The index ofrefraction ofthe glass is 1.50.
512 In an isosceles prism ofprism angle 45°, it is found that when the angle of incidence is same as the prism angle, the emergent ray grazes the emergent surface. Find the refractive index ofthe material ofthe prism. For what angle ofincidence the angle ofdeviation will be minimum ?
Ans. [5cm to the left of the surface of sphere]
Ans. [yjl, 41.5P]
58 A source of light is located at double focal length from a convergent lens. The focal length of the lens is/= 30 cm. At
513 An astronomical telescope with objective offocal length 100 cm and eyepiece of focal length 10 cm is used by a
what distance from the lens should a flat mirror be placed so
that the rays reflected from the mirror are parallel after passing through the lens for the second time ?
shortsighted man whose far point is 33 cm from his eye, to form an image ofan infinitelydistantobject at his far point. Find the separation ofthe lenses, and magnification obtained.
Ans. [45cm]
Ans. [107.5 cm, 13.3]
iGeometricar optics
331
514 Figure5.371 shows a right angled prismy45C having
518 An image Fis formed ofa pointobjectJFby a lenswhose
refractiveindex ="loweredintowater ^j.Findangle
opticaxis is^5 as shown in figure5.372. Draw a ray diagram to locate the lens and its focus. Ifthe image Fofthe objectA'is formed by a concave mirror (having the same optic axis AB) instead of lens, draw another ay diagram to locate the mirror and focus. Write down the steps of construction of the ray
a so that the incident ray normal to face AB will be reflected at face BC completely.
diagrams.
•y
Figure 5.372
Ans. [
Figure 5.371
angles.
519 A thin planoconvex lens.of focal length/is split into two halves : One ofthe halves is shifted along the optical axis (figure5.373)The sqjaration between objectand image planes is 1.8m. The magnification ofthe image formed by one ofthe halflens is 2. Find the focal length ofthe lens and separation between the two halves. Draw the ray diagram for image
Ans. [60° and 40° ]
formation.
Ans. [ a >sm  
515 An equilateral prism deivates a ray through 40° for two incidence angle which differ by 20°. Find the two incidence
516 A ray of light strikes a glass slab ofthickness t. (i) Prove that it emerges on the opposite face, parallel to the initial ray. 1.8m—
(ii) Prove that the value of deflection of beam which passed through the plate is:
Figure 5.373
Ans. [40cm, 60cm] 1sin 2,i1
1
t sin i
«^/sin^'i
(iii) Prove that for a small angle of incidence z,, the internal shift Xis given by
520 The focal lengths of the objective and the eyepieceof a compound microscqje are 1 cm and 5 cm respectively.An object placed at a distance of 1'. 1 cm from the objective has its final image formed at (i) infinity (ii) least distance ofdistinct vision. Find the magnifying power and the distance between the lenses. Least distance of distinct vision is 25 cm.
1
x = ti,
1
Ans. [(i) 16 cm,  50; (ii) 15.17 cm,  60] g J
517 A planoconvex lens has a thickness of 4 cm. When
521 Find the minimum size of mirror required to see the fiill image of a wall behind a man standing at the centre ofroom, where His the height ofwall.
placed on a horizontal table with the curved surface in contact
Ans. [H/3 ]
with it, the apparent depth ofthe bottommost point ofthe lens is found to be 3 cm. Ifthe lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre ofthe plane face of the lens is found to be 25/8 cm. Find the focal length ofthe lens.
placed making 10°withthe mirrorMy Findthepositions ofthe first two images formed by each mirror. Find the total number ofimages.
Ans. [75cm]
Ans. [10° and 50° from A/," and 20° and 40° from M^, 11]
where
is the refractive index ofglass with respect to air.
522 Two mirrors are inclined by an angle 30°. An object is
Geometrical Optfcs
1332
523 AB is a man ofheight 2m and Mis a mirror of length 0.5m and mass 0.1 kg. Initially top ofmirror Mand A are at the same level and the Mstarts falling freely always remaining vertical. If the level ofthe eyes ofthe man is 1.5 cm below his head .^4,
526 See the following figure5.377. Which of the object(s) shown in figure will not form its image in the mirror.
•O4
find the time after which the man sees the reflection ofhis feet.
I
Figure 5.377
M
Ans. [O3]
527 Twoplane mirrors are inclined at an angle of 75® to each other. Find the total number ofimages formed when an object is placed as shown in fIgure5.378.
Figure 5.374 Ans. [0.318s]
524 Figure5.375 shows a point object.^ and a plane mirror MN. Find the position of image of object/I, in mirror MV, by drawing a ray diagram. Indicate the region in which observer's
eye must Be present in order to view the image.
✓
V/////////////////A
Figure 5.378
M
f N
Figure 5.375
Ans. [4]
528 Two plane mirrors are inclined at an angle of70° to each other. Find the total number of images formed when object is placed as shown in figure5.379. Total images = 5
Ans. [ ef
'////////////////////.
525 An object is placed at.^(2,0) and MT^is a plane mirror, as shown in figure5.376. Find the region on Taxis in which reflected rays are present.
iv
P(4.3)
Figure 5.379 Ans. [5]
529 There is a point object placed in front of a plane mirror. If the mirror is displaced 10cm away from the object, find the distance by which its image will get displaced. Ans. [20cml
M
^(4.2)
A
(2, 0)
Figure 5.376 Ans. [Reflected rays exist on Faxis between (0,6) and (0,9)]
530 A crown glass prism of refracting angle 8° is combined with a flint glass prism to obtain deviation without dispersion. If the refractive indices for red and violet rays for the crown glass are 1.514 and 1.524 and for the flint glass are 1.645 and 1.665 respectively, find the angle of flint glass prism and net deviation. Ans. [1.53"]
iGeometrical Optics
333J
531 An opaque cylindrical tank with an open top has a 535 Figure shows a torch producing a straight light beam
diameter ofS.OO mand iscompletely filled with water. When
felling on a plane mirror at an angle of 60° as shown in figure5.383. The reflected beam makes aspot Ponthe vertical sunlight ceases to illuminate any part of the bottom of the screen as shown. If at r = 0, mirror starts rotating clockwise
the setting sun reaches an angle of37° above the horizontal,
tank. How deep is the tank ?
about the hinged with an angular velocity = 1° per second.
Ans. [4 m]
Find the speedof the spot on screen aftertime t = 15s.
532 In the situation shown in figure5.380, find the velocity vector ofimage in the coordinate system shown infigure.
Screen '//////////////////z
5 m/s
fixe
incident
ray
10 m/s
Figure 5.380
Figure 5.383
Ans. [ 5(1 +VJ)/ +5y m/s]
271
Ans. [ — m/s]
533 Find the velocity of the image of a moving object in situation shown in figure5.38I in which object and mirror velocities in horizontal and vertical directions are shown. 5 m/s
536 Alight ray/is incident onaplane mirror M The mirror is rotated in the direction as shown in the figure5.384 by an arrow atthe frequency (Wtt) rev/sec. The light reflected by the mirroris received on the wall IFwhich is at a distance 10m from
i
the axis ofrotation. When the angle ofincidence becomes 37°, find the speedof the light spot on the wall.
O 10 m/s
30 m/s 6 m/s
Figure 5.381
Ans. [70.178 m/s]
534 Two plane mirrors are inclined toeach other atan angle
lOm
W
30°toeach other. Arayoflightis incident at anangle of40°to Figure 5.384
the mirror (M^). Find the total angle ofdeviation ofthe ray after the third successive reflection due to mirrors.
Ans. [lOOOm/s]
537 A sphericallight bulbwith a diameterof 3.0 cm radiates
light equally in all directions, with apower of4.57c W. (a) Find the light intensity atthe surface ofthe bulb, (b) Find the light intensity 7.50 mfrom the centre ofthe bulb, (c) At7.50 m, a convex lens is set up with its axis pointing toward the bulb. The lens has a circular face with a diameter of 15.0 cm and a 40>
Figure 5.382
Ans. [160° Clockwise ]
focal length of30.0 cm. Find thediameter ofthe image ofthe bulb formed ona screen kept at thelocation oftheimage, (d) Findthe light intensity at the image. Ans. [(a) 5000 W/m^; (b) 0.02 W/m^; (c) 0.125 W/m^; (d) 288 W/m^j
G^metrical OpticsJ 
538 A coin liesonthebottom ofa lake2mdeep at a horizontal
542 A small object ofheight 0.5 cm isplaced in front of a
convex surface ofglass (p = 1.5) ofradius ofcurvature 10cm. beam oflight situated 1mabove the surface ofthe liquid of Findthe height ofthe image formed inglass.
distance xfrom thespotlightSwhich isa source ofthin parallel
refractive index p=
as shown in figure. The liquid height is
2m. Findx sothat a narrowbeamoflight fromS whenincident
on the liquid surfece atincidence angle 45° fells directly on the coin.
Ans. [1cm]
543 Infigure5.387 shown AB isaplane mirror oflength 40cm
placed ataheight 40 cm from ground. There is alight source S atapointon the ground. Find the minimum and maximum height ofaman (eye height) required tosee the image ofthe source if he is standing at a point P onground shown in figure. T E O
I.
com
K20cmi
40cm
Figure 5.387
Figure 5.385
Ans. [320 cm]
Ans. [ I 1+
I Im]
544 Aplane mirror ofcircular shape with radius r =20cm is fixed to the ceiling. A bulbis to be placed on the axis of the
539 What should be the value of angle 0 such that light
entering normally through thesurface
ofa prism («=3/2)
does not cross the second refracting surface AB.
mirror. A circular area of radius
= 1 m on the floor is to be
illuminated afterreflection oflightfrom the mirror. Theheight ofthe room is 3m. What should be the maximum distance from the centre ofthe mirror where bulb is to be placed so that the
requiredarea is illuminated? Ans. [75 cm]
545 A room contains air in which the speed of sound is 340 m/s. The walls ofthe room are made of concrete, in which
thespeed ofsound is 1700 m/s. (a) Find the critical angle for total internal reflection of sound at the concreteair boundary,
(b) In which medium should thesound be travel to undergo total internal reflection ? Figure 5.386
Ans. [(a) sin"' ]y ]; (b) air]
Ans. [ 6/3z +x=10 is incident atapoint
Ans. [5 m].
Pon the face/4Rofthe prism.
551 Aray in incident normally on aright angle prism whose refractive index is ^ and prism angle a =30°. After crossing the prism, ray passes through a glass sphere: It strikes the
glass sphere at ^ distance from principal axis, as shown in
(a) ,Find thevalue ofp for which thefay grazes theface /IC
(b) Findthedirectionofthe.mallyrefractedrayifp=
3
(c) Find the equation ofray coming out ofthe prism ifbottom BC is silvered ?
the figure5.389. The sphere is half polished. Find the total angle of deviation of the incident ray after all reflections and refractions from thisoptical setup.
(0, 0, 0)
Figure 5.391
Figure 5.389
Ans. [{a) ^ ; (b) Along z axis; ic)yl3z +x=\0 ]
Ans. [ISO")
555 A convexlens of focal length 20 cm and a concavelens
552 Ashortsighted man, the accommodation ofwhose eye offocal length 10 cm are placed 10 cm apart on the same optic isbetween 12 cm and 60 cm wears spectacles through which axis. Abeam oflight travelling parallel to the optic axis and he can see remoteobjects distinctly. Determine the minimum havinga beamdiameter5.0 mm, is incidenton the convex lens. distance at which the man can read a book through his Showthat the emergent.beam is parallel to the incident one. Findthe beam diameter oftheemergent beam. spectacles. Ans. [15 cm]
Ans. [2.5 mm]
Opticsl i336
556 A thin convex lens of refractive index ^ = 1.5 is placed 559 A glass rod has ends as shown in figure5.394. The
between apoint source oflight Sand a screen A, as shown in refractive index ofglass is p. The object Oisata distance 2R the figure. Light rays from the source Sare brought tofocus on from the surface of larger radius of curvature. The distance the screen A, forming apoint image P.The distance SPisequal between apexes ofends is 3R. Find the distance ofimage formed ofthe point object from right hand vertex. What isthe condition to 50 cm. Water
p=—Iis now poured into avessel interposed on p for formation ofa realimage.
between the object and thelens, and it isobserved thatwhen Glass
the water level is 8 cm the screen has to be moved up by a
distance of6 cmin order togeta sharp image. Findthe focal
"
length ofthe lens. Screen
7:
•2R
Figure 5.394
560 When the object is placed 4 cmfrom the objective of a
microscope, the final image formed coincides with theobject. The final image isatthe least distance ofdistinct vision (24 cm). Ifthemagnifying power ofthe microscope is15, calculate the
Figure 5.392
focal lengths of the objective and eyepiece.
Ans. [12 cm]
Ans. [/•(, = 3.125 cm and/^ = 7.5 cm]
557 Athin equiconvex glass lens (p^ = 1.5) is being placed on the top ofa vessel ofheight /i = 20 cm as shown in the 561 A hemispherical portion ofthesurface ofa solid glass figure5.393. Aluminous point source is being placed atthe sphere (p= 1.5) ofradius r issilvered to make the inner side bottom ofthevessel ontheprincipal axis ofthelens. When the reflecting. An object isplaced on the axis ofthe hemisphere at air is on both the sides of the lens, the image of luminous sourceis formed at a distance of 20 cm from the lens outside the vessel. When the air inside the vesselis being replaced by
a distance,3r from thecentre ofthe sphere. Thelight from the
object is refracted at the unsilvered part, then reflected from the slivered part and again refracted at the unsilvered part.
a liquid ofrefractive index Pj, the image ofthe same source is Locate the final image formed. being formed at a distance,30 cm from the lens outside the
"
vessel. Find the Pj.
'
fi, = 1
T
2r
2()cm
Luminous
Figure 5.395
source
Figure 5.393
Ans. [At the pole of slivered face]
Ans. [1.11]
562 The focal lengths of the objective and the eyepiece of an astronomical telescope are 0.25 m and 0.02m, respectively. 558 Light passes symmetrically through a 60® prism of The telescope is adjusted to view an object at a distance of refractive index 1.54.After emergence out from the prism the 1.5 mfrom theobjective, thefinal image being 0.25 m from the light ray is incident on a plane mirror fixed tothebase ofthe eye ofthe observer. Calculate the length ofthe telescope and
prism extending beyond it. Find the total deviation ofthe light the magnificationproducedby it.
rayafterreflection fromthe mirror.
Ans. [31.85 x lO'^ m, 16.2] Ans. [0®]
•GTOmetrical Optics 337
563 Athin converging lens offocal length/= 1.5 misplaced
569 In a simple astronomical telescope the focal length of
along_vaxis such that its optical centre coincides with the the object glass is0,75 m and thatoftheeyepiece is 0.05 m.
origin. Asmall light source Sisplaced at ( 2.(5 m, 0.1 m) as
Calculate the magnifying power when the final image ofadistant
shown infigure5.396. Where should aplane mirror inclined at object is seen (a) a long way off, (b) at a distance of 0.25 m. an angle 0 with tan 0 = 0.3, be placed such that^coordinates Find the distance between the two lenses in each case.
offinal image is0.3 m. Also find a: coordinate offinal image.
Ans. [(a) 0.0417 m (b) 0.7917 m]
570 An astronomical telescope consisting of two convex lensesof focal length 50 cm and 5 cm is focussed on the moon.
What is the distance between the two lenses in this position? If the telescope is then turned towards an object 10 maway, how much would the eyepiece have to be moved to focus on Figure 5.396
the object without altering the accommodation of the eye? Calculate the angular magnification produced by the telescope in the two adjustments.
Ans. [5m, 4m]
Ans. [10]
564 A ball is kept at a height yQ above the surface of a 571 How would you use two planoconvex lenses of focal index ji. At ^= 0, the ball is dropped to fall normally on the lengths 6cm and 4cm todesign aneyepiece free from chromatic transparent sphere of radius R, made ofmaterial ofrefractive
sphere. Find the speed of the image formed as a function of
time for t
D
2
k where n = 0,1,2 2
# Illustrative Example 6.6
" •
The coherent point sources
and 5*2 vibratingin samephase
emit light ofwavelength k. The separation between the sources
is 2k. Consider a linepassingh through $2 andperpendicular to the line 5,5*2. What is the smallest distance from where a minimum of intensity occurs due to interference ofwaves from the two sources? H
D
d=A.9x 10^m=0.49mm
(6500xl0»)(2)^0 0366mn, 15x10"^
# Illustrative Example 6.9
Change in fringe width
Twoslits in Young's interference experimenthave width in the ratio 1 : 25. Deduce the ratio of intensity at the maxima and
= p1 p = 0.0866  0.065 = 0.0216 mm
minima in interference pattern.
# Illustrative Example 6.11
Solution
In Young's double slitexperiment theslitsare0.5mmapartand
The intensities dueto separateslits are in proportionto the slit width. Theamplitude is proportional to the squareof intensity.
interference is observed on a screen placed at a distance of 100 cm fromthe slits. It is foundthat the 9'*^ brightfringeis at a distance of 8.835 mm from the second dark fringe from the
centre of the fringe pattern. Find the wavelength of light used. 502
Solution
^2 = 50,
{0^+02?
The distance ofn''^bright fringe from the central fringe is given
(a,+5iJ,)^
by
Now
XD
36a{ 16a?
36 _ 9 16 ~ 4
^nuxU = 9:4.
where p = {XDUd)is the fringe width.
For 9^^ bright fringe,
.
X5=9P
...(6.31)
^Wave Optics 353
The distance of dark fringe from the central fringe is given by X '=n
{2n\)U) Ad A = t7sin n —
2
2 d
«lP
For2"'^ dark fringe, Figure 6.19
...(6.32)
In above situation the phase difference between waves from
Fromequations(6.31) and(6.32), weget
slits
and Sj at screen center C is given as
15
,
271
(p= ~
X
X(j = Dsin0
...(6.36)
Above equation(6.36) gives theposition where path difference iszero or the position ofcentral bright fringe after changing the illuminated earlier than 5, sowaves from will lead inphase overthe wavesfrom 5", becauseof the extra path d sin0 which direction ofincident light beam ondouble slit plane. Ifthe parallel light beam incident onthedouble slitplane normally then both the wavefront illuminating slit will travel till itreaches slit 3",. the slits will getilluminated simultaneously and the point of Ifweanalyze thepath difference inthetwo waves interfering at zero pathdifference (central maxima) would be located atpoint screen center C then the two waves travel equal path after slit Candiftheincident beam isrotated downward upto the situation the inclined incidence of beam we can see that slit
will be
plane upto this point but due to initial path difference before
slitplane, the total pathdifference in waves at pointC will be d sin0.
shown in figure6.19 thenwithabove analysis wecan statethat during rotation ofthe beam the interference pattern will shift upward bya distance given byequation(6.35).
Wave Ophcsf .
6.4.2 Effect of Submerging YDSE Setup in a Transparent
Thetimetaken byray1 intravelling through theglass slabis
Medium
pw
w
In a transparent medium ofrefractive index m, when a light of wavelength / enters then we know its wavelength and speed reduces which are given as
c/pj
c
Path length covered byray2 in space while ray1 was travelling in slab is
c V = ~ jl
and
"k X= ~ X
pw
/ = CX —
.,.(6.38)
= pw
c
Ifa YDSE setup is submerged ina transparent liquid asshown in figure6.20 then in the submerged part due to decreased Thus path difference between thetwo rays after theglass slab wavelength fringe width of the interference pattern also is given as decreases which is given as
d
A= ptv w= w(p 1) ...(6.37)
ixd
... (6.39)
If these two light waves (rays) are brought to focus of a converging lens as showi then the two waves will interfere with a phase difference given as
Fringe Pattern in Screen
Shrunken fringe pattern in water
Narrow slits
27C
...(6.40)
Water
Ifeachofthe lightwave in the two thin light beams (rays) have
Water tank
intensity 1q then at the focal point ofthe lens the resulting intensity of light is given as
Incident light beam
Figure 6.20
,, ' • f7tw(pl) /, =4/,cos2^ =4/,cos^
...(6.41)
6.4.4 EffectofPlacinga Thin Transparent Film in front ofone of the slits in YDSE Setup
Here p, sowecansaythat on submerging a given YDSE setup ina transparent medium, theinterference pattern shrinks. 6.43 Path difference between twoparallel waves due to a denser medium in path ofonebeam
Figure6.22 shows a YDSE setup and in front ofslit5, a thin transparent film ofthickness t and refractive index p isplaced. Duetothisfilmthe lightcoming outfrom slit will getdelayed
bysome path asitgets slower inthe film medium. Atthe screen center where thephysical pathdifference iszerointhetwolight
Figure6.21 shows two coherent light rays (thin beams) from a waves coming from slitiSj and S^, due tothin film an optical single source of light travelling in same direction parallel to path difference isintroduced inthepath oflight waves which is each other. Ifin pathoffirst beam a glass slabofrefractive index
p is placed which is ofwidth wthen the light ray1 will slow
givenbyequation(6.39).
down after it enters in slab at point Aand its speedwill reduce to c/m.Whentheray1 whichentersin slabcomes outat point B then in this duration the ray2 which was travelling in space would have travelled a longer path as it was travelling at speed c. C A= ?(fi 1)
Screen
Figure 6.21
Figure 6.22
[Wave Optics
355
Optical path difference in the twolightwaves from the slitsat
this case obviously it is not necessary that at screen center
screen center C is given as
there isa bright fringe because the path difference at point Cis now given as A^=/(p  1). Ifwe find the ^zvalue'' at screen
A = /(pl)
/ ...(6.42)
center then it is given by equation(i) as
Atscreen center theextra path given by above equation(a) is
introduced in the light coming from slit .Sj as it got delayed
%l)
while travelling inside thefilm. If we tryto locate a point on screen where optical path difference iszero then we can clearly
X
state that it will be locatedabovepoint C on screen where the
...(6.47)
Saytheexpression obtained inequation(6.47) gives a numerical value 6.23 aftersubstituting the numerical values ofallconstants
extra optical path introduced in first wave is compensated by init.Thatmeans atpoint Conscreen thepath difference inthe the extra physicalpath travelled by the second wavewhich is two waves is 6.23X or it shows that the original center bright given by equation(6.42). If P is the point of net zero path fringe is located about 6 fringes above thepoint Cwhere path difference located at a distance Xq from thescreen center C in difference will be zero.
figure6.13 then at point
we can use
Thus at any point on screen if you calculate the 'zvalue' then
dXr,
it directly gives you anidea about thepath difference ofbright ...(6.43)
Above equation(6.43) gives the shiftof interference pattern on screen due to insertion of thin film in front ofone of the slits
or dark fringes above or below this point which helps us in calculating the total number of fringes in any region of interference pattern.
HIllustrative Example 6.12
becausethe fringe of zero path differencewas locatedat screen
center inYDSE before the film was inserted infront of5,.At a A double slit arrangement produces interference fringes for general pointPon screen located in upper half of screen at a sodium light {X = 5890 A) that are 0.40° apart. What is the angularfringe separation ifthe entirearrangement is immersed
distancex from C, the path differenceis given as
in water ? dx
...(6.44) Solution
And at a general point P' located in lower half of screen at a
distance x from C, the path difference is given as
In double slit arrangement, the angularseparation is given by
dx
XD
...(6.45)
6.4.5 Concept of zvalue in Interference Pattern of YDSE
At any point in interferencepattern on YDSE screen, we can define a numeric parameter 'z' for relating pathdifference inthe two light waves and wavelength oflight as
2d
Let angular separation in air and water be respectively. Now
2d
andp^=
X^,D
'W
"W
Ap=Zpk
,(6.48)
P.
Ao
...(6.46)
We know that
Above numerical parameter Zp is a constant which gives the multiplier of lightwavelength at anypointonscreen that gives the path difference at a the specific point on screen. This
or
Velocity of light in air Velocity of light in water
Va
f^A
'W
fXw
'zva/we' helpsin quicklydetermining thenumberofbrightand dark fringes in any length of interference pattern of YDSE.
where/is the frequency oflight.
For example ifwe look at the YDSE Setupshown in figure6.14
Now
inwhich a thin film isplaced in front ofslit.Sj due towhich the overall fringe pattern is shifted upward by some distance. In
and
From equation(6.48) and (6.49), we have
(6.49)
Wave Optics [
f356
hL
A=i/=3.4x ICHmx — K
P. A=
or
3.4x10^ ,
—^=54SAX
6.2x10"
Total no ofmaxima on x aixs
by D are= 548
Illustrative Example 6.13 # Illustrative Example 6.15
A YDSE setup is immersed in water (p = 1.33). If has slit separation 1 mm and distance between slits and screen is 1.33m.
Incident lightonslits have wavelength 6300A. Findthefringe width on screen.
In YDSE setup slits are illuminated by a light of wavelength
4000A and a lightofunknown wavelength. It is observed that fourth dark fringe ofknown wavelength coincide with second bright fringe of unknown wavelength. Find the unknown wavelength.
Solution
W/L oflight in water
Solution
X
6300 ,
X = —= A ® p 1.33
For
dark fringe from centre {Nl)XD
In YDSEfringe width is given as
^NDX„D
6300x10"'® X1.53
d
1.33x10"^
A =
N=4,X=moA
for
7x4xlO~^xD
= 6.3xia^m
^4D
=0.63 mm
For
2d
bright fringe from centre NXD
^IllustrativeExample 6.14
Thereare two coherent sources 5", and
2d
^NDwhich produce waves
in same phase placed on Y axis at points (0, 2d) and (0, d)diS shown in figure6.23. A detector Displaced at origin which moves along X direction, find the number of maxima recorded by detector excluding points x = 0 and a:= co. Given that wavelength of light produced is 6200 A and separation
^
N=2,X = 1
for
2XD
d Given that
AD
7x4xlO"^D
between sources is 0.34 mm.
2^ =>
2XD
~ d A.= 7x 107m= 7oooA
# Illustrative Example 6.16
In a YDSE setup, light of wavelength 5000 A is used and slit separation is 3 x 10"' m. When a transparent sheet ofthickness
1.5 X10"' m is placed overone oftheslits which has p= 1.17,
A = 548X 0
at CO
A = 548.4?.
A = 0
Figure 6.23
Solution
Path different between wavesfrom 5", and S^ at origin
find the shift of fringe pattern. Take separation between slits and screen is 1 m.
Solution
At Cpath different A = r(p1)
!Wave Optics 357
sides ofC. At these points violet colour will be missing from
•/'A=0
white light due to destructive interference and the colourseen
lt_4 CA = X(nl)
on screen will be reddish white (not red as it iswhite  violet) because other then violet all other colours will be present at this point asonly violet will be completely absent.
Bluish
Figure 6.24
At point P we have
Reddish A= —
dx
^
Reddish
^ 1x1.5x10"^ xQ.17 a
3x10"^
Bluish
Double slit plane
a: = 0.055 m = 8.5 cm Screen
Illustrative Example 6.17
Figure 6.26
In YDSE setup find the distance between two slits that result
Similarly ifwemove fiirther away from point Conscreen next
inthe third minimum for 4200 Aviolet light atanangle 30®.
point will be the point ofdestructive interference ofred light which ishaving the maximum wavelength inwhite light where
Take d«D. Solution
At point P on screen path different between waves from 5", and
is A = Srl2
III min
j^2 ••m D
red colour will beabsent and path difference atthis point will be /2. As the wavelength ofred colour isapproximately double that ofviolet colour soat this point violet color light will also have higher intensity and the colour offringe here will appear ^Bluish' (Not blue asother colours are also present here). Beyond this point onscreen allcolours will mix and nospecific colour
will be seen on screen and screen looks 'Mixed White' or 'Off White'mcoXour.
6.4.7 Effect ofChanging SUt Width inYDSE Setup
'Ot/sin 0
In a YDSE setup if each slitisproducing light intensityon screen then this intensityis directlyproportional to the width Figure 6.25
Asin 30®
d X
5X
 ~:r (forII min) 2,
£/=5X=5x4200x lO'Om = 2.1x10^ m
6.4.6 Use of White Light in YDSE
When white light is used in YDSE setup, all the component waves of white light will have zero path difference at screen
ofthe slits indouble slitplane. Figure6.27 show that a plane parallel beam oflight ofintensity/, illuminates thedouble slit plane. Ifeach slit isofwidth wand length / then the light power which pass through these slitswill be given as ...(6.50) ^siit h • Thus thepower oflightfrom each slitwill beproportional tothe
slit area given by equation(6.50) and aswe have already studied insection ofspherical and cylindrical waves that intensity due to any light sourceis givenbythe ratio ofsourcepowerand the surface area of wavefront at any point. We know that a slit produces halfcylindrical wavefronts after double slitplane and
center Candallcolors inwhite light will intafere constructively and a bright white fringe is produced. But.as we move away
the wavefront W^ soata distancex from a slit thelight intensity
from the center then closest to screen c^ter first the destructive
can be given as
interference of violet color will take place as violet color has minimum wavelength in white light at a point shown in
figure6.26 where path difference will become XyH on both
hlit
^Slit
LJw
Tixl
•Kxl
/iw
= ~
...(6.51)
Wave Optics
i358
Thusthe lightintensityon screen at a distance D from double slit plane due to each slit is given as /jW,
...(6.52)
4lit~
Thus the light intensity by each slit on screen is directly proportional tothe slitwidth andinversely proportional to the separation between double slit plane and the screen.
N
^
Intensity
/ i AU ; Central Bright Fringe D.
h \ A'v'V'V
N
\
\
\ Screen
Intensity /,
f' / ' VV
D, B,
i i
0 Central Bright Fringe D,
1 1 1 •"

'
Figure 6.28
6.4.8 Fresnel's Biprism as a Limiting case of YDSE
/'/ // / ' // y
/
Fresnel demonstrated interference pattern similar to YDSE by
using a biprism. The biprism consists of two prisms of very Screen
smallrefractinganglesjoined baseto base.It ismade bygrinding
a thin glassplatewithopposite inclination onthetwohalves of the plate on one side as shown in frgure6.29. Figure 6.27 Thin glass plate
Ifboth slits 5, and S2 are ofequal width then intensity byboth slits will be same /q on screen and the resulting intensity of bright and dark fringes will be
Shaded part is removed
and
!
From equation(6.54) we can see that dark fringes will be perfectlyblack and separatingthe bright fringes and in such a case proper interference pattern can be seen in which we can clearly distinguish bright and dark fringes as shown in
Biprism Perspective view
(a)
figure6.27.
Ifthe width ofthe slits are unequal as shown in figure6.28 then
the intensities of the twoslits on screenwillbe different, say/^,
and /q2 respective then the light intensity ofbright and dark fringes on screen will be (b)
hrighl ^
^01 ^02 ^4^01^02 •••(655)
^Dark ~(4^ ^ 4^ ~^01 ^02 ~^4^01^2 •••(656) From above equation(6.56) we can see that the intensity of dark fringe is not zero due to which in the interference patter we cannot clearly distinguish bright and dark fringes as shown in figure6.28.
Figure 6.29
We've already studied that a light ray from a source S when incident on a small angled prism at near normal incident as shown in figure6.29(b), it gets deviated by angle ofdeviation 5, and from the other side it appears to be coming from the image ofthe light source 5". The angle ofdeviation 8 is given as 5=.I(pl)
...(6.57)
wave Optics
359'[
the slit and screen is taken as D = + Zj so the fringe width on the screen can be given in the same wave we use for YDSE setup. XD
Fringe width in Fresnel Biprism setup is p=jSubstituting the values ofD and d in above expression we get X(a + b) P = 2aA(^l)
...(6.59)
Figure 6.30
6.4.9 Lloyd's Mirror as a limiting case of YDSE
Now if we look at figure6.30 in which two light rays from a source S incident on a biprism then these bends after refraction through biprism by same angle of deviation toward base of prisms such that the refracted beams appear to be coming from the two images of the source S' and S". Figure6.31 shows the interference setup of'Fresnel'sBiprism ExperimenC in which there is only one slit kept at a distance 'a' fromthe biprism and at a distance 'Zj' fromthe biprism screen is kept. Due to refraction, biprism splits the incident light in two beams and from the other side the light beams appear to be coming from the two images S' and S" of the slit and these beams interfere and producesthe interferencepattern similar to
In 1834Lloydmade a setup to demonstrate interferenceusing a single mirror and a slit of light at almost grazing incidence of light.Figure6.32 shows theLloyd's Mirrorsetupofinterference
inwhich light from theslit5j falls onthemirror andthe reflected light interfere with the direct light on the screen. The reflected light appears to be coming from the image of the slit as shown.For theregion ofscreen in whichthetwo lightsinterfere, interference patternis obtained whichis similarto YDSEsetup.
Region of Interference
YDSE on screen.
X
Screen
Figure 6.32
In the above setup the separation between the two slits is taken
diSd=2h and the separation between slit plane and screen is given as D. If we consider a general point P on the screen in the region of interference then the path difference at point P between the lights from the two sources is given as Figure 6.31
dx A — D
X
...(6.60)
1— 2
In above figure we can see that each light fiom slit S falling on biprism gets deviated by the deviation angle given by equation(6.57). For the region betweenslit plane and biprims
In above equation(6.60), the first term is the physical path difference in the path travelledbythe two light waves reaching at point P and the second term is the extra path introduced in
for the ASOS'wq can use for small value of5, we have
the reflected light due to its reflection from a denser medium at mirror.
— =crtan5i ad 2
In this setup at point P, a bright fringe will be obtained if the
path difference is an integral multiple of the light wavelength so
~ =aA(H'l)
we use
d = 2aA (pl)
...(6.58)
Equation(6.58) gives the separation between the two coherent sources S' and S" from which the interference pattern is being produced and the separation between the plane of images of
dx
X
A=— + r =NX D
2
x.. = {2N\)
...(6.61)
XD 2d
...(6.62)
Wave OpticsJ
J360
Similarly we can say that at point P there will be dark fringe if the path difference in the two waves at P is an odd multiple of half of the light wavelength so we use dx
X
X
shown in figure6.34. If the angle ofconvergence between the light beams is 0 then the light beams after interferenceproduces a fringe pattern as shown in figure where the fringe width is given as
...(6.63)
A= —+ =(27V+ 1)D 1 ^ ^2
...(6.65)
NXD
...(6.64)
^ND
From equations(6.62) and (6.64) we can analyze that the interference pattern obtained in this setup is inverse to that of the YDSE setup. At the location of bright fnnges in YDSE, here we are getting dark fringes and vice verse. But we can see that the distance between two successive bright or dark fringes which is the fringe width remain same. Fringe Width in interference pattern obtained in Lloyd's Mirror setup is given as
XD
XD
d ~ 2h 6.4.10 Billet Split Lens as a limiting case of YDSE
This is an experimental setup shown in figure6.33. Athin lens of small aperture is cut in two halves at the center and these halves are separated by some distance from original principal axis oflens. These halflenses produce two images of the slit as shown in figure. The light from these two images will produce interference pattern on screen in the region where the two images overlap.
Figure 6.34
The derivation of above equation(6.65) is left for students as an exercise.
# Illustrative Example 6.18 Consider the situation shown in figure6.35. The two slits
and ^2 placed symmetrically around the central line are illuminated by a monochromatic light of wavelength X. The separation between the slits is d. The light transmitted by the slits falls on a screen placed at a distance D from the slits. The slit
is at the central line and the slit
is at a distance z
from Sy Another screen £"2 is placeda further distance D away from £,. Findtheratioofthemaximum tominimum intensity observed on £2 if ^ is equal to : XD
(a) ^
XD
(b) V
XD
(c)
4d
Screen
Figure 6.33
This setup is similar to YDSE setup if we look at the plane of two images of the slit producing interference pattern on the screen. By using lens formula and magnification we can find
out the location ofthese images and their separation by which using the formula offringe width in YDSE setup we can find the fringe width in this setup.
Figure 6.35 Solution
6.4.11 Interference of Two Converging Coherent Parallel Beams of Light
When two monochromatic and coherent parallel light beams incident on a surface (or screen) in a converging manner as
Light from sources and get interfered and thereafter and becomes new sources. At the path difference between
the lightscoming from and£2 iszero. Therefore theyinterfere constructively and so iij = (a + fl) = 2a
^Wave Optics 361
(a) At
the path difference Ax = cisinQ = dtan 0
ax —
D
D Screen
Figure 6.36
The path difference
Corresponding phase difference = n radian
^x = S^S^P
«4= 0
("i+atf _(2a +af
The ratio
^min
2a~0
= = 1
= X. cos 0
The maximumpath difference can be
XD dz
(b)
D
cos 0
A
= X
= X,; when cos 0 = 1 or 0 = 0°
•*max
and minimumpath difference
Corresponding phase difference = 2k radian. A_ =0; when cos 0 = Oor 0 = 90®
a^ = a + a = 2a
Now
and
Thus in between these two positions there is only one miima
(2a + 2ay
L
for which
{2a2a)^ Ax = —. Thus •d dz
(0)
L4J — = X cos 0
D
Corresponding phase difference ^ — radian
or
cos 0 = — 2
0 =60®
4^ = 0^ + 0^ + 2aa cos — = 2d^
a Illustrative Example 6,20
 yfla
or
/„i„
(2a +^af _ (3.414)^
In a modified Young's double slitexperiment, a monochromatic
(2a^f
uniform and parallel beam oflight ofwavelength 6000A and
(0,586)^
=34
# Illustrative Example 6.19
intensity ^ W/m^ is incidentnormallyon two circular apertune Aand Bofradii 0.001 mand 0.002 mrespectively. Aperfect transparent film of thickness2000 A and refractiveindex 1.5for
thewavelength of 6000 A is placed infront ofaperture Aas
The two coherent sources of monochromatic light of shown in figure6.37. wavelength Xare located at a separation X. The two sources
are placed on a horizontal line and screen is placed perpendicular to the linejoining the sources. Findposition of the farthest minima from the centre of the sources. Solution
Suppose at P the farthest minima will occur. Let it subtends an angle 0 at the centre of the sources.
Figure 6.37
Wave Optics
[362 _
Calculate theintensity oflightreceived atthefocal point ofthe lens in watt.The lens.is symmetrically placedwith respect to
Screen
1
X
the aperture. Assume that 10% ofthe power received by each aperture goes in the original direction and is brought to the
d
D
focal point.
Figure 6.38 Solution
Solution
The intensities oflight from the sources
andS2 are given as
(a) Thepath difference at O is given as
X71(0.01)2= 105 IT
Ac=
7t;
2D
Forthe darkfringe at 0, this pathdifference should be 10
k = — xjt(0.02)2=4xia5lf
X 3X
Ax=,y,...
V. ^
The intensities ofsources after emerging from the lenses are
j^inimum value ofd, we use
/ A =0.10 X105 lf=l(r^ If
2ylD^+d^2D 
/„ = 0.10x4x 105lf=4x 10"^^ D
The path difference produced dueto film
or
Ax =(^l)r=(1.5l)x2000xlO>o=10^m ]2 ^
271
271
(b=
^
X
=
, 7
X
1+D'
D
or
1/2
,n xlO
^ = 4
6000x10"'®
(})= — radian
Asrefraction through lensdonotintorduce anypathdifference soabove willbethe netphasedifference in thetwolightbeams
D
or
1+ 
4
2D'
D+
or
d^
X
D =
2D
4
superposing atfocal point ofthelense. Sonetintensityreceived at F is given as
d =
or
=10*^ +4X 10^+ 2Vl0"^x4xl0^ cosj
(b) Attheabove calculated value old, first bright fringe will be obtained at a position where the path difference between the two waves will be X.
= 7x106 If
# Illustrative Example 6.21
5,1
Consider the arrangement shown in figure6.38. The distance D is large comparedto the separation ^/between the slits.
(a) Find theminimum value ofd sothatthere is a darkfringe atO.
(b) Suppose ^/has this value. Findthe distance x at whichthe next bright fringe is formed.
(c) Find the fringe width.
5: 4
D
O*
]
0=±2.25>
/p= 7qCOs2 0
...(6.111)
Fromthe above equation(6.111) wecanstate'"When apolarized light ispassed throughan analyzer then the intensity of light
Ifweconsider an oscillation amplitude Ashown bya bold arrow in above figure whichis at an angle0 tothe transmission axisof the Polaroid then its component A cos0 will pass and its component Asin0 willbe blocked. The resulting amplitude of
Wave Optics'
•38,8
thepolarized lightafter transmission through thepolaroid will be sum ofall A cos0 for all the oscillations ofthe field vector in
unpolarized light withvalue of0 varying from 0to2n.Ifintensity
^^foRotaton,
ofunpolarized light is Iq then theintensity ofthe transmitted
Substance
\/
light can be given as
1
Polarizer
Ij.= k ^ ^COS0 e=o
Figure 6.107
As at all angles 0 in unpolarized light, the amplitude of oscillation can be considered same with equal probability so we can write the above expression as
Ij~ (Iq cos' /, =/o(cos
There are two ways in which optically active substances are divided based on the direction of rotation of plane of
polarization. When the emerging light is observed from the average from 0=0 lo 2n
average from 6=0 to 2n
As average valueofsquareof cosine of an anglevarying from
0to 2jc is given as ^cos^ 0^ =—. ...(6.112)
It /()
With the result obtained in above equation(6.112) we can state
"The intensity of unpolarized light after passing through a polarizer reduces to halfinpolarized light". 6.8.9 Optical Activity of Substances
directionoppositeto the directionof propagationof light then the substances which rotates the plane of polarization in clockwise direction (toward right) are called 'Dextrorotatory Substances' and those which rotate the plane ofpolarization of
light in anticlockwise direction (toward left) are called 'Laevorotatory Substances'.
The optical activity of a substance is analyticallymeasured in terms of 'SpecificRotation' ofa substance whichis defined for a given wavelength of light as the rotation produced by one decimeter long column of the solution containing unit concentration (1 gm/cm^) ofthe substance. Ifa polarized light passes througha solution ofoptically active substance having concentration C(in gm/cm^) and oflength / (indecimeter) and the angle bywhich planeofpolarizationof lightgets rotatedby 0 then specificrotation of the substanceis given as Qt—
Some specific transparent substances have property to rotate
the planeof polarization of a polarized light when the light is passedthrough it, suchsubstances are called''OpticallyActive'' substances and this ability of substances to rotate the plane of
polarization of light is called 'Optical Activity'. Quartz and Cinnabar are common examples of opticallyactive crystalsand substances which are opticallyactive in both solid and liquid state are Sugarand Camphor. Figure6.107 shows an optically active substance on which a polarized light is incident with the field vectoroscillations in a plane which is in the plane of paper which is the plane of polarization. We can see in figure, as the light passes through the substance, the plane of polarization rotates and the light which comes out of the material remain
IC
UIllustrative Example 6.31
The axes of a polarizer and an analyzer are oriented at 30° to each other.
(a) If unpolarized light of intensityf is incident on them, what is the intensity ofthe transmitted light?
(b) Polarized lightof intensity/q is incident onthispolarizeranalyzer system. If the amplitude ofthe light makes an angle of 30° with the axis of the polarizer, what is the intensity of the transmitted light ?
linearly polarized but its plane ofpolarization gets rotated. Solution
0Rotatoiy Substance !3«~rfCi
Polarizer
'Stance
(a) Half the light passes through the polarizer, so the intensity of the polarized light incident on the analyser is
1=
/(j, andthe intensity/'ofthe lightpassingh through the
analyzer is/'=/cos^0= —/qCOS^30° = 0.375/q.
[Wave_Optics
389
(b) After passing through the polarizer the intensity is # Illustrative Example 6.33 1=/pCos^ ®~^ cos^ 30° =0.75 7^. This light is now polarized at 30° to the axis of the analyzer, so the intensity of the light The axes ofa polarizer and an analyzer are oriented at right passingh through the analyzer is
/'= /cos^ 0=0.75 /q cos2 30° =0.563/o.
angles toeach other. Athird Polaroid sheet isplaced between them with its axis at45° tothe axes ofthe polarizer and analyzer. (a) If unpolarized light of intensity 7^ is incident on this
UIllustrative Example 6.32
system, what is the intensity ofthe transmitted light? (b) What is the intensity of the transmitted light when the
A beam ofplane polarised light falls normally ona polariser
middle Polaroid sheet is removed ?
(crosssectional area3 x lO^m^) which rotates aboutthe axis
ofthe ray with an angular velocity of31.4 rad/s. Find the energy
Solution
of lightpassing through the polariser per revolution and the intensity oftheemergent beam ifflux ofenergy oftheincident (a) The light that passes through the polarizer has an intensity ray is 10"^ W.
of
and is polarized at 45° to the middle sheet. Thus the
light that passes through the middle sheet has the intensity
Solution
7= 7qCos^ 45° =0.257q and ispolarized 45° tothe axis ofthe If 7q is the intensity of plane polarised light incident on the polariser, then intensityofemerging light is given by
analyzer. Thus the intensity ofthe light passing through the
analyzer is 7'=7cos^ 45° =0.1257^.
7=/qCos^0 The average value ofI over one revolution can be calculated
(b) Ifthe middle Polaroid sheetis removed, wehavea crossed polarizeranalyzer and no lightgetsthrough.
as:
j 2n Web Reference at www.phvsicsgalaxv.com
I = — r 7^/0
 2nl
;Age Group  High School Phj^sies ] Age 1719 Years Section  OPTICS
Topic  Polarization of Light
1
Im = — f 7o cos^ 9 J "
I av
Module Number  1 to 13
Practice Exercise6.4
2
Intensity is given by
(i)
ona second Polaroid whose planeofvibration makes an angle
Power
^0
(a) Ordinary light incident on one Polaroid sheet falls
of 30° with that ofthe first Polaroid. Ifthe Polaroids are assumed
area
tobe ideal, whatis the fraction ofthe original lighttransmitted 10"
L = 3x10
through both Polaroids?
=y10 W/m2
(b)
If the second Polaroid is rotted until the transmitted
intensity is 10percentofthe incident intensity, whatis the new and
,= ^ =1W/ml
angle ? [(a) 3/8; (b) 63.4°]
The energyoflight passingthroughthe polariserper revolution 271
E = IxA^T=I av
5
x^x — CO
271
= _,(3,10)x — = 10^J
(ii) Three nicolsprisms are placed such that, first and third are mutually perpendicular. Unpolarised light is incident on first nicoTs prism, the intensity of light emerges from third nicol's prism is 1/16 the intensity of incident light. Find the angle between first and second nicol's prisms. [22.5°]
Wave •{
;390
Polarized light of intensity 7q is incident on a pair of transmitted intensity is maximum. Thenobtain thetotal intensity ofthe transmitted beam in terms of/q. incidentamplitude and the axes of the first and secondsheet, respectively. Show that theintensity ofthe transmitted lightis
(iii)
Polaroid sheets. Let 0j and 02 be the angles between the
1~1q cos^ 0j cos^ (0j  Sj). (iv) A mixture ofplanepolarised andunpolarised lightfalls normallyon a polarisingsheet.On rotatingthe polarisingsheet about the direction of the incident beam, the transmitted
intensity variesbya factor 4. Find the ratioofthe intensities
and Iq respectively of the polarised and unpolarised components in the incident beam. Nextthe axis ofpolarising sheet is fixed at an angle of 45® with the direction when the
Advance Illustrations Videos at www.phvsicsgalaxv.com
AgeGroup  Advance Illustrations Section  OPTICS
"TopicWave Optics
Illustrations  22 Indepth Illustrations Videos
[Wave Optics
~
"""
"T" """agr
Discussion Question Q61 Why does an aeroplane flying at a great altitude not Q611 Some motor cars have additional yellow headlights, case a shadow on the earth?
Why?
Q62 When sun rays pass through a small hole in the foliage Q612 Why does acolumn ofsmoke rising above the roofof atthetopofahightree,theyproduceaneIIipticalspotoflight houses seem blue against the dark background of the on the ground. Explain why. When will the spot be a circle? surrounding objects, and yellow or even reddish against the background of a bright sky?
Q63 If a mirror reversesright and left,whydoesn't it reverse
up and down?
Q613 What are primary colours ?Why are they socalled ?
Q64 Is itpossible to photograph a virtual image?
Q614 Abeam ofwhite light passing through ahollowprism gives no spectrum. Is this true of false ?
Q65 Is is possible for a given lensto act as a converging lens
inone medium, and asa diverging lens inanother?
Q615 Ordinary paper becomes transparent when it isoiled. Explain whey.
Q66 A cameralensis marked//1.8. Whatis the meaning of
this mark?
Q616 Whey does the moon, purely white during the day, have a yellowish hue after sunset ?
Q67 If these are scratches on the lens of a camera, they do
notappearonaphotographtaken with the camera. Explain. Do
Q617 What is the best position of the eye for viewiing an
the scratches affect the photograph at all ?
object through a microscope?
Q68 The sun seems to rise before itactuallyrises and itseems
Q618 What is the function ofthe circular stop at the focal
tosetlong after it actually sets. Explain why.
plane ofthe objective of a telescope?
Q69 Does the focal length ofalens depend on the medium in Q619 Why isthe objective ofatelescope oflarge focal length which the lensis immersed? Is it possible for a lensto act as a
andlargeaperture ?
conversing lens in one medium and a diverging lens in another
medium?
Q620 Can the optical length between two points even be less than the geometrical path between these points ?
Q610 Explainwhy the use ofgogglesenablesan underwater' swimmter to see clearly under the surface ofa lake.
*
^
^
Wave Optics
:392
ConceptualMCQs Single Option Correct 61 A parallelmonochromatic beamoflightis incidentnormally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction ofincident beam. At the first minima of the diffraction pattern, the phase difference between the rays coming from the edges of the slit is : (A) 0
(B)f
(C) TT
(D) 2sz
62 In Fraunhofer diffraction experiment, L is the distance between screen and the obstacle, b is the size of obstacle and "k is wavelength ofincident light, the general condition for the applicability ofFraunhofer diffraction is: (A)
— »1 LX
(B)
^ =1
(Q
^«i Lk
P)
^.1
Lk
Lk
63 In a Fraunhofer diffraction experiment at a single slit using a light of wavelength 400 nm, the first minimum is formed at an angle of30°. The direction 0 of the first secondary maximum is given by:
• 1^2
(A) sin .(Q sin
1
(B) sinM^ (D) sin
(A) tan ' Ij
(B) tan' ! 
(Q tan 1
P) tan"' I
4^
67 Which ofthe following cannot be polarized ? (A) Ultraviolet rays P) Ultrasonic waves (Q Xrays P) Radiowaves
68 An unpolarised beam of intensity 7^ falls on a polariod. The intensity ofthe emergent light is :
(A) ^
P) ^0
(Q f
p)' Zero
69 Which ofthe following is a dichroic crystal ? (A) Quartz P) Tourmaline (Q Mica p) Selenite
610 An unpolarised beam ofintensity /qis incident ona pair ofNicol's prisms making an angle of60° with each other. The intensity oflight emerging from the pair is:
(A) 7, '
P) 7^2
(Q V4
P) 7^/8
611 When anunpolarized light ofintensity 7^ isincident on a polarizing sheet, the intensity ofthe light which does not get transmitted is:
64 If we observe the single slit Fraunhofer diffraction with wavelength k and slit width b, the width of the central maxima is 20. On decreasing the slit width for the same k: (A) 0 increases P) 0 remains unchanged (C) 0 decreases p) 0 increases or decreases depending on the intensity of light 65 In diffraction from a single slit, the angular width of the central maxima does not depend on :
(A) P) (Q P)
Xof light used Width ofslit Distance of slits from screen Ratio ofX. and slit width
(A)
P)
(Q Zero
P) ^
612 An optically active compound: (A) Rotates the plane polarised light P) Changing the direction ofpolarised light
(Q Do not allow plane polarised light to pass through P) None of the above
613 A 20 cm length ofa certain solution causes right handed rotation of38°. A30 cm length ofanother solution causes left handed rotation of24°. The optical rotation caused by 30 cm length ofa mixture of the above solutions in the volume ratio 1 :2 is:
66 The critical angle of a certain medium is sin~' polarizing angle ofthe medium is:
(A) P) (Q P)
Left handed rotation of 14° Right handed rotation of 14° Left handed rotation of3° Right handed rotation of 3°
_Wave Optics 393
614 Specific rotation ofsugar solution is 0.5 deg m^/kg. 200 kgm3 of impure sugar solution is taken in a smaple polarimeter tube oflength 20cm andoptical rotation isfound tobe 19®. The percentage ofpurity ofsugar is : (A) 20% (Q 95%
(B) 80% p) 89%

615 Two polaroids are placed in the path ofunpolarized beam
CA)f ^I
bd
(Q
2d
^ ^ Id' Ad
Id
620 Ratio ofintensities between a point.4 and that ofcentral
fringe is 0.853. Ifwe consider slits are ofequal width then path
of intensity such that no light is emitted from the second Polaroid. If a thirdpolaroid whose polarization axismakes an
difference between two waves at point Awill be:
angle 0 with the polarization axis offirst polaroid, is placed
(A) I
(B) T
(Q V
(D)X
between these polaroids, then the intensity oflight emerging from the last polaroidwill be: (A)
— sin^20 cos'^0
(Q
2
(B)
^sin^20 4
p) L ens'* 0
621 Two waves of same intensity produce interference. If intensityatmaximum is4/,then intensity atthe minimum will be: (A) 0 (Q 3/
p) 2/ p) 4/
616 Two light waves arrives ata certain point on a screen.
The waves have the same wavelength. At the arrival point,
622 Intensity ofcentral bright fringe due to interference of
their amplitudes and phase differences are: (I) 2aQ, 6aQand7irad (n) 3oq, 5c7gand7iand (HI) 9aQ, la^, and3^ rad (IV) 2aQ, 2aQ and0.
two identical coherent monochromatic sources isI. Ifoneofthe
The pair/s which has greatest intensity is/are: (A) I • P) H (C) II, m P) I,IV
source is switched off, then intensity ofcentral bright fringe becomes:
(A) I
P)/
617 Two beams oflight having intensities/and 4/interfere to producea fringepatternonthe screen. Phasedifference between
(B)
623 In an experiment to demonstrate interference offlight using
7U thebeams is —at point Aandtc at point B. Thedifference in Young's slits, separation of two slits is doubled. In order to maintain same spacing of fringes, distance D of screen from intensities ofresulting light at points AandB is :
(A) 3/ (Q 5/
P)4/ p)6/
slits must be changed to : (A) D
(B) f
618 A plane electromagnetic wave offrequency Wq falls normally on the surface of a mirror approaching with a
(Q 2D
relativisitic velocity v. Then frequency of the reflected wave
willbe ^given p=— j: (A)
IzP
U+P
624 In Young's double slit experiment, 12 fringes are obtained tobe formed in a certain segment ofthe screen when light of wavelength 6000A is used. Ifwavelength oflight is changed to 1+ P
Wn
screen is given by:
(l + P)wo (Q
(1p)
4000A, number offringes observed in the same segment ofthe
(D)
(1P)
(A) 18
p) 24
(Q 30
p) 36
(1+P) Wo
619 White light isused to illuminate two slits inYoung's double
625 In Young's double slit experiment, interference pattern is found to have an intensity ratio between bright and dark fringes
slit experiment. Separation betweenthe slits is b and the screen
as 9. Then amplitude ratio will be:
isata distance d(» b) from the slits. Then wavelengths missing
(A) 1 (Q 9
at a point on the screen directly in front ofone ofthe slit are:
P) 2 P) 3
•
Wave Optics
^3^
626 IfA is the amplitude ofthe waves coming from a point
TL
source at distance r then which of the following is correct:
(A) cos 0= ^
(B) cos 0  —
(A) Accr^ {Q Acct^
(Q sec 0  cos 0 =
p) sec 0  cos 0 =
(B) ^ cc r' (D)^ccr'
4?.
627 If^ istheamplitude ofthewave coming from alinesource at a distance r, then :
(A)^ocr
(B)v4xH'2
{Q A
(P) Ax
633 In ayoung double slitexperiment, D equals thedistance ofscreen and is the separation between the slit. Thedistance ofthenearest pointtothecentral maximum where theintensity is same as that due to a single slit, is equal to :
^28 Phase difference between two coherent light waves (A) ^DX having same amplitude A is Itz/S. If these waves superpose each other at a point, then resultant amplitude at thepoint of
DX
2DX
DX
P)
superposition will be:
(A) 2A
(B) 0
(Q A
(D) A^
629 Ratio ofamplitudes ofthewaves coming from two slits
634 Inthegiven figure6.109, ifa parallel beam ofwhite light is incident on the plane of the slits then the distance of the white spot onthescreen from Ois [Assume dv>wv.physicsgalaxy.coni 61 Aprism having angle A° (which isvery small) isplaced in front of a point source 5 at a small distance d. A screen is
placedat a large distanceDas shownin the figure6.113. Find the fringe width of interference pattern given that source is emitting light of wavelength Xand refractive index of prism
of100 cm from theslit. Itisfound thatthe9^ bright fringe isat a distance of 7.5 mm from the second dark fringe from the center ofthefringe pattern on same side. Findthewavelength of the light used. Ans. [5000 A]
IS p.
H—H •D
Figure 6.113
I
66 Light ofwavelength 520 nmpassing through a double slit, produces interference pattern of relative intensity versus deflection angle 9 as shown in the figure6.115. Find the separation
between the slits.
Ans. [ 7—• , ] ^ iy\)Ad '
62 A thin glass plateof thickness t and refractive index p is inserted between screen & one of the slits in a Young's experiment. If the intensity at the center ofthe screen is /, what was the intensity at the same point prior to the introductionof
0(degrees) Figure 6.115
Ans. [1.99 X 10"^ mm]
the sheet.
67 The distance between two slits in a YDSE apparatus is
Ans. [/„ = / sec^ \
3mm. The distance ofthe screen from the slits is Im. Microwaves
of wavelength I mm are incident on the plane of the slits 63 A young's doubleslit arrangement produces interference fringes for sodium light(X = 5890A) thatare0.20® apart. What is the angular fringe separation if the entire arrangement is immersed in water(refractive indexofwater is4/3). Ans. [0.15°]
normally. Find the distance of the first maxima on the screen from the central maxima. Ans. [35.35cm app., 5]
68 One radio transmitter.^ operating at 60.0 MHz is 10.0m
from another similar transmitter Bthatis180° outofphase with 64 Two coherent narrow slitsemitting lightofwavelength X transmitter How far must an observer move from transmitter in the same phase are placed parallel to each other at a small Atowards transmitter B along the lineconnecting AandB to separation of 2X. The light is calculated on a screen 5 which is reach the nearest point wherethe two beams are in phase ?
placed a distance (D » X) from the slit 5", as shown in figure6.114. Find the distance x suchthat the intensity at P is equal to the intensity at point O.
Ans. [1.25m]
69 Thecentral fringe oftheinterference pattern produced by the light ofwavelength 6000 Ais found toshift totheposition of4^^ bright fringe aftera glass sheet ofrefractive index 1.5 is introduced. Find the thickness of the glass sheet. Ans. [4.8 pm]
'D
Figure 6.114
Ans. [JiD]
65 In Young's doubled slit experiment the slits are 0.5 mm apart and the interference is observed on a screen at a distance
610 A board source oflightofwavelength 680nm illuminates normally two glass plates 120 mm long that meet at one end and are separated by a wire 0.048 mm in diameter at the other
end. Find the number ofbright fringes formedover the 120mm distance. Ans. [141]
I40'2 \
:
"
.
611 Two microwavecoherent point sources emitting waves
^
^ "l
'Waye^Optlcs.1
ofrefraction index \iisfilled: (wavelength oflight inairisXdue
ofwavelength Xare placed at5X distance apart. The interference to the source, assume l»d,D » d). isbeing observed on a flat nonreflecting surface along a line
passing through one source, in a direction perpendicular to the line joining the second source (refer figure6.116) Considering as4mm, calculate the positions ofmaxima and draw shape ofinterference pattern. Take initial phase difference
H/s
between the two sources to be zero. Observation surface Screen
Figure 6.118
(a) Between the screen and the slits.
Figure 6.116
(b) Between the slits & the source 5. In this case, find the Ans. [48
minimum distance between the pointsonthe screen wherethe intensity ishalfthemaximum intensity onthescreen.
, 21, 4r,o, m.m; (((©)))]
612 In a YDSE with visible monochromatic light two thin
transparent sheets are used in front of the slits
Ans. [(a) Acj) =
"I / dJ X '
2 2d '•
and S'j.
p, =1.6 and Pj = ^•4 Ifboth sheets have thickness t, the central 615 In aYDSE, aparallel beam oflight ofwavelength 6000 A maximum is observed at a distance of5 mm from center O. Now is incident onslits at angle ofincidence 30°. A&Bare two thin the sheets are replaced by two sheets of same material of transparent films, each ofrefractive index 1.5. Thickness of^ is refractive index
t =
h+h
but having thickness
suchthat
. Now central maximum is observed at a distance of
2
20.4 pm. Light coming through A&Bhave intensities I&41 respectively on the screen. Intensity at point O which is symmetric relative to the slits is 31. The central maxima is above O.
8mm from center O on the same side as before. Find the
thickness/,(inpm) [Given: i/= 1mm.D= 1m].
A
ll
^ 30^ y
0.1mm
1
/15
Im
(u,, / > n, tj)
Figure 6.119 Figure 6.117 Ans. [33]
(a) What istheHfiaximum thickness of5to do so ?Assuming thickness of B to be that found in part (a) answer the following parts:
613 Inatwo slitexperiment with monochromatic light, fringes are obtained ona screen placedat somedistance from the slits. If the screen is moved by 5 x m towards the slits, the change in fringe width is 3 x 10""^ If the distance between the slits is 10"^ m, calculate the wavelengthof the light used.
(b) Findfiinge width, maximum intensity&minimum intensity
Ans. [6000 A]
(d) Intensityat 5 cm on either sideof O.
614 A sourceS is kept directlybehindthe slit
in a double
slitapparatus. What will bethephase difference atP, ifa liquid
on screen.
(c) Distance ofnearestminimafrom O. Ans. [(a) t, = 120 nm; (b) [J  6 mm; (d) I (at 5 cm above 0) = 9/]
= 91,
 /; (c) p/6 = 1 mm
[VVave^Optics
403]
616 A central portion with a width ofd= 0.5 mm is cut out of a convergent lens having a focal length of 20 cm. Both halves
same distance from theslitbyplacing a convex lensat a distance
of0.3 mfrom the slit. The images are found tobe separated by
are tightly fitted against each other and a point source of 0.7 X10~^ m. Calculate the wavelength ofthelight used. monochromatic light(A = 2500A) isplaced in front ofthe lens Ans. [5550 A] at a distance of 10cm. Findthe maximum possible numberof interference bands that can be observed on the screen. Ans. [5]
621 Two identical monochromatic lightsources A andB of
intensity 10"'^ W/m^ produce wavelength oflight 4000>/3 A.
A glass ofthickness 3 mm is placed in thepath oftheray as 617 A plastic film with index ofrefraction 1.80 isputonthe shown in the figure6.121. The glasshas a variablerefractive surface ofa carwindow toincrease thereflectivity and thereby index n=l +Jx wherex (in mm) is distance ofplate from left to keep the interior of the car cooler. The window glass has to right. Calculate resultant intensityat focal point F ofthe
index ofrefraction 1.60.
lens.
(a) What minimum thickness isrequired, iflight ofwavelength 600 nm in air reflected from the two sides of the film is to
interfere constructively?
(b) Itisfound tobedifficult tomanufacture andinstall coatings as thin as calculated in part (a). What is the next greatest thickness for which there will also be constructive interference? Ans. [(a) 8.33 x lO""^ m; (b) 2.5 x lO"'] Figure 6.121"
618 Two coherent monochromatic sources AandB emit light
Ans. [4 X io'5 W/m^]
ofwavelength X.The distance between A and B is d= 4X. \D
622 Two fine slits are placedvery close to each other and theyare illuminated bya strong beam of light ofwavelength 5900 A. Fringes are obtained at a distance of 0.3 m from the slits. The fringe width is found to be 5.9 x 10"^ m. Calculate the distance between the slits. Ans. [3 X 10"^ m]
Figure 6.120
(a) If a lightdetector is moved alonga line CDparalleltoAB, what is the maximum number ofminima observed ?
(b) If the detector is moved along a line BEperpendicular to AB and passing through B, what is the number of maxima observed ?
623 Biprism fringes are produced with the help of sodium light (A = 5893 A). Theangle ofthe biprism is 179.5? and its refractive index 1.5. If the distance between the slit and the
biprism is 0.4mandthe distance between thebiprism andthe screen is 0.6 m, find the distance between successive bright fringes. Also calculate the maximum width ofthe slit for which the fringes will still be sharp. Ans. [3.376 x 10^ m, 112 ^m]
Ans. [(a) 8, (b) 4]
624 A glass plate 1.2 x 10~^ m thick isplaced in thepath of 619 A beam of lightconsisting oftwo wavelengths, 6500 A
and 5200 Ais used to obtain interference fringes in aYoung's double slit experiment. Findthedistance ofthethirdfringe on thescreen from thecentral maximum for thewavelength 6500A. What is the least distance from the central maximum at which
one of the interfering beams in a biprismarrangement using monochromatic lightofwavelength 6000 A. If the central band shifts by a distance equal to the width of the bands, find the refractive index ofglass. Ans. [1.5]
the bright fringes due to both wavelengths coincide? The distance between the slits is 2 mm and the distance between
the plane ofthe slits and the screen is 120 cm. Ans. [1.17 mm, 1.56 mm]
625 A monochromatic light of intensity 7 is incident on a glass plate. Another identical glass plate is kept close to it. Each glass plate reflects 25 per cent of the light incident on it and transmits the rest, find the ratio of the maximum and
620 In an experiment using Fresnel biprism, fringes ofwidth 0.185 X 10"^ m are observed at a distance of 1 m from the slit. Images are of the coherent sources are then produced at the
minimum intensities in the interference patternformed bythe twobeam obtainedafter reflection from eachplate. Ans. [49 : 1]
Wave Optics j
1404
626 Aplane light wave ofwavelength X=700 nmfalls normally on the base of a biprism made of glass (p = 1.560) width refracting angle0 = 5°.Aplateisplaced behind thebiprism and thespace between themisfilled witha liquidofrefractive index (p'= 1.500). Find thefiinge width onascreen behind thebiprism.
631 Two pointcoherent sources ared= 4X apart. Two light detectors are moved  one along the line parallel to the line
joiningthe sources andtheotheralong thelineperpendicular to this line and passingthrough one of the sources. Find the maximum number ofmaxima registered by each detector. Ans. [9, 9]
632 In aYoung's arrangement, the distance between the slits is 1.5 mm and the distance ofthe screen from the slits is 2m. At
what minimum distance from the central fringe will red and
violet brightfringes coincide iftheslitis illuminated withwhite light?The wavelength range in white light is from 420 nm to 690 nm.
Ans. [12.88 mm] Figure 6.122 Ans. [66.8 pm]
627 In a biprism experiment, the angular width of fringes was 0 = 2'. calculate the separation between the slits if the wavelength of the incident light was X= 700 nm. Ans. [1.2 mm]
633 A doubleslit apparatus is immersed in a liquid of refractiveindex 1.33.It has slit separationof 1mm, and distance between the plane of slits and screen is 1.33 m. The slits are illuminatedbya parallelbeamof light whosewavelength in air is 6300 A.
(a) Calculate the fringewidth, (b) one of the slits of the apparatus is covered by thin glass sheet of refractive index 1.53. Find the smallest thickness ofthe sheet to bring adjacent
628 In a biprism experiment, calculatethe intensity oflights on the screen at a point 5p/6 away from the central fringe in termsofthe intensity ofthe central fnnge,which \sIq. Here P is
minimum on the axis.
the fringe width.
634 In a Young's double slit experiment, the separation
Ans. [0.633 mm, 1.58 mm]
between the slits is 2 x 10~^m whereas the distance of screen
Ans. [3/^/4]
from the slits is 2.5 m. A light of wavelengths in the range of
629 Twoparallel beams of light P and Q (separation d and 2000 A  8000 A is allowed to fall on the slits. Find the mutuallycoherent)are incidentnormallyon a prismas shown, wavelength in the visible region that will be present on the ifthe intensitiesofthe upper and the lowerbeams, immediately screen at 10"^ m from the central maximum. Also find the after transmission from the face AC, are 4/and /respectively, wavelength that will be present at that point of the screen in find the resultant intensity at the focusofthe lens which brings the infrared as well as in the ultraviolet region. Ans. [8000 A, 4000 A 2666.7 A ... 8000 A (infrared), 2666.7 A
them into focus.
(ultraviolet)] p
635 A narrow slit ofwidth 0.025 mm is illuminatedby a parallel
\
beam of light. The diffraction pattern is observed through a telescope. It is found that to reach the first minimum, the telescope has toberotated through 1°24'from thedirection of the direct ray. Calculatethe wavelength of light used.
d
, i Q
Figure 6.123
Ans. [6100 A] Ans. [91]
630 Twopoint sources are d=nk apart. A screen is held at right angles to the line joining the two sources at a distance D from the nearer source. Calculate the distance of the point on the screen where the first bright fringe is observed. Assume D»d. 2D(D + nX) Ans. [J—^ ^1
636 Angular width of central maximum in the Fraunhofer diffractioh pattern ofa slit is measured.The slit is illuminated by another wavelength, the angular width decreases by 30%. Calculate the wavelength of this light. The same decrease in angular width ofcentral maximum isobtained when theoriginal apparatus is immersed in a liquid. Findthe refractive indexof the liquid. Ans. [4200 A, p = 1.43]
Optics
405 i
637 In a Young's double slit set up the wavelength oflight
641 In a Young experiment the light source is at distance /j = 20pm and /j=40pmfrom the slits. The light ofwavelength
used is 546 nm. the distance ofscreen from slits is Im. The slit separation is 0.3 mm.
X=5000A is incident on slits separated atadistance 10 pm. A
(a) Comparethe intensityat a pointP distant 10mm fromthe
screen is placed at a distance D = 2m awayfrom the slits as
central fringe where theintensity is1^.
shown in figure6.125. Find
(b) Findthe numberof brightfringes between P and thecentral fringe.
.Ans. [(a) 3.0 X106 m; (5) 3 q x lO"' 7^]
638 In a double slit pattern (X. = 6000A), the firstorderand tenth order maxima fall at 12.50 mm and 14.75 mm from a
particularreference point. If Xis changedto 5500A, findthe position of zero order and tenth order fringes, other arrangements remaining the same. Ans. [ 12.25 mm, 14.55 mm]
Figure 6.125
639 An interference is observed due to two coherent sources
(a) The values of0 relative to the central line where maxima appear on the screen ?
placed at originand ^2 placed at (0, 3X, 0). Here Xis the wavelength of the sources. A detector D is moved along the (b) Howmany maximawillappearon the screen? positive xaxis. Find xcoordinates on thexaxis (excluding (c) What should be minimum thickness ofa slab ofrefractive X= 0 andx = 00) wheremaximumintensityis observed. index 1.5beplaced onthe pathofoneoftheraysothatminima occurs at C?
Ans. [1.25X, 4?.]
(b) 40; (c) 5000A] 640 AYoung double slitapparatus isimmersed ina liquid of Ans. [(a) sin"' refractive index p,. Theslitplane touches theliquid surface. A parallel beam ofmonochromatic light ofwavelength X(inair) in 642 A screen is placed 50 cm from a single slit, which is
incident normally on the slits.
illuminated with 6000 A light. Ifdistance between the first and third minima in the diffraction pattern is 3.0 mm,what is the width ofthe slit? Ans. [ 0.2 mm ]
643 In a double slit experimentthe distancebetween the slits is 5.0 mm and the slits are 1.0 m from the screen. Two interference
patterns can be seen on the screen one due to light with
wavelength 480nm, andtheother duetolight with wavelength 600 nm. What is the separation on the screenbetween the third Figure 6.124
(a) Find the fringe width
orderbright fringes of the two interference patterns? Ans. [0.072 mm]
(b) If one oftheslits (say 5'2) is covered by a transparent slab ofrefractive index ^2 and thickness (as shown, find the new 644 Ina Young's double slitexperiment using monochromatic position ofcentral maxima.
light the fringe pattern shifts by a certain distance on the screen
(c) Now the other slit 5, is also covered by a slab of same thickness and refractive index pj asshown infigure6.124 due
when a mica sheet ofrefractive index 1.6 and thickness 1.964
microns is introduced in the path of one of the interfering
to which the central maxima recovers its position find the value
waves. The mica sheet is then removed and the distance
between the slits and screen is doubled. It is found that the
ofp3.
(d) Find theratio ofintensities at Ointhethree conditions (a), (b) and (c),
distance between successive maxima(or minima) now is the same as observed fringe shift upon the introduction of the
mica sheet. Calculate the wavelength of the monochromatic (b)
(c)
(d)
light used in the experiment. Ans. [589 nm]
• Wave Optics j
i406
645 Two very narrow slits arespaced 1.80 pmapart and are reflection theintensity decreases by90% andoneach refraction placed 35.0 cm from a screen. What isthe distance between the theintensity decreases by10%, find theratio oftheintensities first and second dark lines of the interference pattern when the slits are illuminated with coherent light of ^ = 550 nm? (Hint: The angle 0 is not small)
ofmaximum to minimum in reflected pattern.
Ans. [12.6 cm]
646 In a Young's double slit set up the wavelength of light used is 546 nm. The distance of screen from slits is 1 m. The slit separationis 0.3 mm.
(a) Compare the intensity ata point P distant 10 mm from the
Figure 6.126 Ans. [361]
centralfringewherethe intensity is/q.
651 A convergent lenswitha focal length of/= 10cmis cut
(b) Find the number ofbright fHnges between P and the central
into two halves that are then moved apart to a distance of d= 0.5 mm (a double lens).Findthe flingewidthon screenat a distance of 60 cm behind the lens if a point source of
fringe. Ans. [(a) 3.0 X lO"^ /(,; (b) 5]
647 Interference pattern with Young's double slits 1.5 mm
apart are formed on ascreen ata distance 1.5 mfrom theplane ofslits. In the path ofthe beamof one ofthe slits, a transparent film of lOmicron thickness and of refractive index 1.6 is
interposed while in thepath ofthebeam from the other slita transparent film of15 micron thickness and the refractive index
monochromatic light {X = 5000 A)isplaced infront ofthelens at a distance ofa = 15 cm from it. Ans. [0.1 mm]
652 In theYoung's double slit experiment a pointsource of X= 5000 Aisplaced slightly offthecentral axis asshown inthe figure6.I27.
1.2is interposed. Find the displacement of the fringepattern. Ans. [3 mm]
648 A glass plate (« = 1.53) that is 0.485 pm thick and surrounded byairisilluminated bya beam ofwhite lightnormal
T
1 . 10mm T Inim *
tP
5mm
to the plate.
(a) What wavelengths (inair)within thelimits ofthevisible spectrum {X = 400 to 700 nm) are intensified inthereflected
m—nr—
2m
beam?
Figure 6.127
(b) What wavelengths within the visible.spectrum are Ans. [(a) 424 nm, 594 nm; (b) 495 nm]
(a) Find the nature and order ofthe interference atthe point P. (b) Findthenature andorder ofthe' interference at O. (c) Where should we place a film ofrefractive index p= 1.5
649 An oil film covers the surface of a small pond. The
and what should be its thickness so that maxima ofzero order
refractiveindex of the oil is greater than that of water.At one
is obtained at O.
point on thefilm, thefilm hasthesmallest nonzero thickness
Ans. [(a) 70"' order maxima; (b) 20"' order maxima;
for which there will be destructive interference in the reflected
(c) t  20 pm, in front of 5,]
intensified in the transmitted light?
light when infrared radiation with wavelength 800nmisincident normal to the film. When this film is viewed at normal incidence
653 A convex lens offocallength/= 25 cmwascut alongthe
at this same point, for what visible wavelengths, if any, will
diameter into two identical halves and was again set leaving a
there be constructive interference? (Visible light has
thin gap of width a = 0.10 cm. A narrow slit, emitting
wavelengths between 4000A and 7000A) Ans. [5330A]
650 A ray of light is incidenton the left vertical face of the
glass slab. Ifthe incident light hasan intensity / andoneach
monochromatic lightofwavelength X= 0.60pmwasplaced at the focus of the lens. A screen was placed behind the lens at a distance b = 50 cm. Find the width ofthe fringes on the screen and the possible number of fringes. Ans. [1.5 X 10^ m, 13]
jWave Optics
407 !
654 Light ofwavelength X—500 nm falls on two narrow slits placed a distance = 50 x 10^ cm apart, at an angle (j) = 30°
658 Inthe YDSE the monochromatic source ofwavelength X ^
relative to the slits as shown in figure6.12S. On the lower slit a is placed atadistance —from the central axis (as shown in the transparent slab ofthickness 0.1 mm and refractive index —is
figure6.129), where is the separation between the two slits
and Sy
placed. The interference pattern is observed on a distance Z) = 2m from the slits. Then calculate:
Imm
i..
"77???777/Z"
^
Mirror
^
H—Scm
D. = 2m
190cm
Figure 6.129
(a) Find the position ofthe central maxima (b) Find the order ofinterference formed at O.
(c) Now S is placed on centre dotted line. Find the minimum thickness of the film ofrefractive index p.= 1.5 to be placed D
Figure 6.128
(a) Position ofthe central maxima? (b) The order ofmaxima at point Cofscreen? (c) How many fringes will pass C, ifwe remove the transparent slab from the lower slit?
Ans. [(a) At 0 = 30° below C; (b) 50; (c) 100]
655 A doubleslit arrangementproducesinterference fringes forsodium light(A, = 5890A) thatare 0.20° apart. Whatwillbe the angular fringe separation if the entire arrangement is 4' immersed in water
p =
3
infront ofS2 sothatintensity at Obecomes —th ofthemaximum intensity. TakeA, = 6000A;t/=6mm Ans. [(a) 4 mm above 0; (b) 20; (c) 2000 A]
659 Fringes are producedbya Fresnel'sbiprism in focalplane ofa reading microscope which is 100 cm from the slit. A lens inserted between the biprism and the eyepiece gives two images of the slit in two positions of the lens. In one case the twoimages ofthe slit are 0.45 mm apart in the othocase2.90 mm
apart. If sodium light of wavelength 5893 A used, find the width ofthe interference fringes. If the distance between the slit and biprism is 10 cm and refractive index ofthe material of the biprism is 1.5, calculate the angle in degrees which the inclined faces ofthe biprism makes with its base.
Ans. [0.15°] Ans. [2.91 or 1.73 mm]
656 In a doubleslit pattern(A, = 6000A), the zeroorder and tenthorder maxima fall at 12.34 mm and 14.73 mm from a
particularreference point. If A, is changed to 5000A, findthe position of the zero order and tenth order fringes, other arrangements remaining the same. Ans. [Zeroorder at the same position but tenthorder maxima at 14.53 mm]
660 In a biprism experiment withsodium light(A. = 5893A), the micrometer reading is 2.32 mm when the eyepiece is placed at a distance of 100 cm from the source. Ifthe distance between
two virtual sources is 2 cm, find the new reading ofmicrometer when the eyepiece is moved such that 20 fringes cross the field ofview. Ans. [2°]
657 The distance between a slit and a biprism ofacute angle 2° is 10 cm. Find the fringe width and the width of the entire band when observation is made on a screen at a distance of
90 cm from the biprism. The refractive index ofthe material of
thebiprism is 1.5 andA, = 5890A. Ans. [0.168 mm, 31.5 mm]
661 In two slit experiment with monochrmoatic light, fringes are obtained on screen placed at some distance from the slits. If the screen is moved by 5 x m towards the slits, the
change in fringewidth is 3 x 10"^ m. If the distance between the slits is 10"^m, calculate the wavelength of the light used. Ans. [6000 A]
Wave Optics]
408
662 In a Young's double slit experiment, the separation between slits is 2 x 10"^ m whereas the distance ofscreen from
by another sheet of thickness = 1.25 pm as shown in figure6.131.Both sheets are made of same material
the slits is 2.5 m. A light of wavelengths in the range of 2000  8000 A is allowed to fall on the slits. Find the wavelength
inthevisible region thatwill bepresent onthescreen at 10"^ m from the central maxima. Also find the wavelength that will be
S, L.
present at that point of screen in the infrared as well as in the ultraviolet region. Ans. [2000 A (ultraviolet)]
663 A double slit apparatusis immersedin a liquidofrefractive index 1.33.It has slit separation of 1mm, and distance between the plane of slits and screen is 1.33m. The slit is illuminated by
M. \*
Figure 6.131
aparallel beam oflight whose wavelength inair is6300 A. (a) Calculate the fringewidth
(b) One of the slits ofthe apparatus is covered by a thin glass sheet of refractive index 1.53.Find the smallest thickness of
the sheet to bring the adjacent minimum ofthe axis. Ans. [(a) 0.63 mm; (b) 1.575 pitn]
664 The Young'sdouble slit experiment is done in a medium
ofrefractive index 4/3. Alight ofbOOOAnm wavelength isfalling on the slits having 0.45 mm separation. The lower slit ^2 is coveredbya thin glass sheetof thickness 10.4 pm and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown in figure6.130 "
havingrefractiveindex p = 1.40.Waterin filledin spacebetween diaphragm and screen. A monochromatic light beam of
wavelength A= 5000Ais incident normally on thediaphragm. Assuming intensity of beam to be uniform and slits of equal width, calculate ratio ofintensity at C to maximum intensity of interference pattern obtained on the screen, where C is foot of
perpendicular bisector of 5, 52. Refractive index of water
Ans. [] 666 In a Young's experiment, the upper slit is covered by a thin glass plate of refractive index 1.4 while the lower slit is covered by another glass plate, having the same thickness as the first one but having refractive index 1.7. Interference pattern
is observed using lightofwavelength 5400 A. It is found that
5
,
Screen
the point P on the screenwherethe central maximum (n = 0) fall beforethe glass plates were inserted now has (3/4) the original intensity. It is fiarther observed that what used to be the fifth maximum earlier,lies belowthe pointP while the sixth maximum lies above P. Calculate the thickness of the glass plate. (Absorption of light by glass plate may be neglected).
Figure 6.130
(a) Find the location of the central maximum (bright fringe with zero path difference) on theyaxis.
Ans. [9.3 X 10"^ m]
(b) Find the light intensity at point O relative to the maximum fringe intensity.
667 In Young's experiment, the source is red light of
(c) Now, if 6000A nm lightis replaced bywhitelightofrange 4000 to 7000A, find thewavelength ofthe light that form maxima exactly at point O. [All wavelengths in this problem are for the given medium of refractive index 4/3. Ignore dispersion] Ads. [(a) 4.33 mm; (b) 0.75
(c) 1300A, 650A, 443.34A, 260A]
665 A screen is at a distance £) = 80 cm from a diaphragm
having twonarrow slits 5"] and
whichare 2 mm apart. Slit5,
is covered by a transparent sheet of thickness
= 2.5 pm and
wavelength 7x10"^ m. When a thin glass plate ofrefractive index 1.5 at wavelength is put in the path of one of the
interfering beams,the central bright fringe shifts by 10~^m to the position previouslyoccupiedbythe 5^^ bright fringe. Find the thickness ofthe plate. When the source is now changed to
green lightofwavelength 5 X10"^m, the central fringe shifts to a position initiallyoccupied bythe 6^^ bright fringedueto red light. Find the refractive index of glass for green light. Also estimate the change in fringe width due to the change in wavelength. Ans. [1.6, 5.7 x IQ^ m]
jAtomic Physics
409
ANSWER & SOLUTIONS CONCEPTUAL MCQS Single Option Correct
1 4 7 10 13 16 19 22 25 28 31
(D) (B) (A) (A) (A) (B) (A) (C) (C) (C) (A)
2
(D) (A) (D)
5 8
11 14
17 20 23 26
.29
3 6 9
(B) (D) (D) (C) (A) (B) (B)
12 15 18 21 24 27 30
(C) (B) (A)
(C) (D) (A) (B) (A) (B) (B)
'Be
IIL
Be
^Be
z
"Be = ^
(iii) As
cc —— we use n
E
Be
= 4
NUMERICAL MCQS Single Option Correct 1 4
(C) (B)
7 10 13 16
(A) (B) (B) (D)
19 22 25 28 31 34 37 40 43 46 49
(D) (A) (B) (A) (C) (B) (D) (A) (D) (A) (A)
(1)^
(A) (D) (B) (B) (C) (B) (D) (C) (C) (C) (D) (D) (B) (B)(A) CD)
2 5 8
11 14 17 20
23 26 29 32 35 38
41 44 47
(D) (C) 9 (D) 12 (C) 15 (B) 18 (B) 21 (B) 24 (D) 27 (C) 30 (B) 33 (C) 36 (B) 39 (D) 42 (A) 45 . (C) 48 (C) 3 6
(iv) E — 3.4eVisfor« = 2stateandweuse 1
V
cc —
"
n
V
''2=2
E^ cc — we use
(v) As
n
±u_ n,;
H
ADVANCE MCQs One or More Option Correct 1 1 4 7 10 13 16 19 22 25 28
(D) (All) (A, C, D) (A. D) (A, C, D) (A) (B) (A, B, C) (D) (B) (A, B, D)
2 2 5 8 11
14 •
17 20
(D) (B, C)
3
(All) (A, C) (B. C) (A, B, C) (B, C, D) (C, D)
6
3
9
12 15 18 21
(A, C, D) (A, C) (A, D)
23
26 29
24 27 30
(C) (A, D) (A. D) (B, C) (A, C) (B) (B, D) (A, B) (A) (A, C) (B, D)
Solutions ofERA CTICE EXERCISE 1.1
"irli] x(3)' «i/=3
(vQ As
TE^
(vii)As
=> (i)
As
KE. =
2r„
TE =
KZe^ 2r^
=l
KE^ =~TE^
XE=+3.4''eV'
oc
Solutions ofPRACTICEEXERCISE 1.2 (i) Minimumwavelength is correspondingto transition from
'lO
/•io = 100 X1.06=106A
00 —> 1 hence
1 ^ 12431 „ (ii) As
cc —
we use
^
13.6
Atomic PhySfcSi
andmaximumwavelength is fortransition2 12431 10.2
1
A
When six wavelength are emitted that means electrons are excited tow= 4and byabsorption of10.2 eV ifelectrons excite from to «2 ~ 4 we use 1__J_
10.2 = 13.6 Z2
^max
2
16
V"i
4
y
for Z= 1 this is not possible
for Z = 2 we get n^ = 2 (ii) We use
X=
^ =4113.2A
for Z = 3 onward this is not possible
Thus Z= 2andtransition is from n^ = 2Xon^ = A. Minimum wavelength (maximum energy) emission is for transition 4
(iii) (a) We use
.
12431
'^min
Maximum wavelength emission isfor transition 4)•3
(b) From« = 3 totalspectral linesare
We use
. =77^^ 12.75x4 =243.74A
= 114.26
• 13.6(3)2
1
=3
A£: =13.622 [ii Z2=P?^=36 5x13.6
12431
^max 0.66x4 =4708.71 A. Solutions of PRACTICEEXERCISE1.3
(i) Frequency if e~ revolution in fn =
Z=6
A:£'in;i=l is A:£',=+13.6Z2 = 489.6eV
X=t{^^=25.39A 13.6x36
and
orbit is
47i^K^Z^e^m n'h'
for transition from «to « 1 orbit radiation frequency is given by
A£ (v) We use =>
12431
12431
10.2
2.55Zi2 Z=2
v =
Energies offirstfour levels ofX are
£:j=_13.6x4=54.4eV £:^=_3.4x4= 13.6eV £'3=1.51 x4= 6.04eV £^=0.85 x4= 3.4eV Iffor thisatom is+E^ = 54.4 eV (vi) Given that and
E^Ej = \0.2 +17.0 =27.2 eV E^E^= 4.25 +5.95=10.2 eV E3£i=27.210.2=17eV
(«l) 2T^K^Z^e'm h' •
1
1
2Ti^K^Z^e^m
v =
n'
Zn\
rp{n\f
for large n wecan use2«1 = 2« and « 1 = « Asi^K^Z'^e^m v =
^fn
3?.3
n'h
(ii) First lineofBalmer series ofHatom is
'Kh
U
9j
rn^+ntfi
first line ofBalmer series for Tatom is
1.89Z^ = 17eV Z=3
4
9J
rrig+mj
1 J_
Now we use
272 =13.6Z2 472 Xj.
36(Wg 5RmH
« = 6
12431
(vii) Incident photon energy.£= ^213 ~ 10.2 eV
)
36(We +mj) SRmj
36
(m^ + nifj )mp  {m^ + mj)mu
5R
mffmj
IAtomic Physics
"
(h) For inelastic collision Hatom must at least exite from n = 1 36
to « = 2 byobsorbing 10.2eV energy. In givensituation initial momentum is zerothusmaximum possible energyloss can be
5R
36
9.109x10"^'x2xl.67xl0"^'^
5x10967800
3X(1.67x10"")
2X^mv^ so we use wv2 = 10.2eV
=2.387 A
10.2x1.6x10"'^
1.67X10"
(ill) Photon
(v) (a) We use mvr = r—
for « = 1
2;c
h
Wavelength of incident photon is "12431 1=
Photon momentum is
v =
2nmr
= 2266.28 A
5.4852
mv
and
m( h ^_ _
P=
^
= evB
X
r \2Tzmr} ^
By conservation of energy we use
AE= i m^vf+
(1)
2neB
By conservation ofmomentum we use
...(2) and
h
j^=W^.V2C0S
Yih
^
(b) Usingmvr= — and
...(3)
= evB weget
nh r =
2K.eB
Squaring adding (2) and (3) we get
h
(c)
heB
nh
2jcot
V27cw^k
2;ce5
1 2 2 ^^
r
1 2Wg
2 2 h^
— myf= —
Vis maximum when n = 1
_
From equation(l) we use .rr
heB
.,2
AE= z—V, 
2m„
2mJX
2
wl"
A£' + 
2m.V = 14.2 m/s
Vn = m
•Li
+ 1 me
By angular momentum conservation we use
from equation(3) we use
mvb = mv^
...(1)
by energy conservation we use cos 9 =
Xwy^V2
0 = 88.9°
= 0.0178
1 2 :rwv/+ 1 2^ rm\p= 2
2
^
1
ZZ'e'
An e.
Sabtitutingvalueofv^from (1) to (2) we get
...(2)
Atomic PhysioS
;4i2
1
n. 1
( v6 1 ,
4mv2 fl4^ for
Sol. 5 (A) In Rydberg's formula the wave numberis directly
t.2
1
proportional to Z^ and Balmer series of hydrogen atom is corresponding to the transition of electron from any level to
ZZ'e
n = 2. Shortestwave length for the spectrumof hydrogenlike
r.2
1
ZZ'e
atom for transition to « = 4 level will be inversely proportional to t? and ifit matches with that ofthe Balmer series ofhydrogen
4n Gq
atom then we use
6 = 0 we get 1 All'
2ZZ'e^ =0
mv
Z2=  x42 = 4
(vii) Usingforce on massm is conservative field F =
dr
Z=2
 mb^r
for circular orbit ofparticle mv
mb^r =
...(1)
r
nh
and
mvr =
(2)
2%
Sol;7 (A)' It has already beendiscussed in theorythat by the wavenature ofelectron and itsDeBroglie wavelength weanalyze for formation of stationary waves in an orbit the orbit circumference shouldbe an integralmultipleofthe wavelength so for consistancy of quantization postulate option (A) is
m(br)r=^ nh r =
Sol. 6 (B) Balmer series in thehydrogen atom haswavelength of most of its lines lying in the visible region of the electromagnetic specturm and someare in infraredregion.
correct.
2%mb
Solutions ofCONCEPTUAL MCQSSingle Option Correct
Sol. 8 (D) Transition A is correspondingto infinity to « = 1 whichis the last line (serieslimit) ofLyman series,transitionB
Sol. 1 (D) As energyin
Orbitof hydrogen atom is givenas
is from « = 5 to « = 2 which is third line of Balmer series and transition C is from « = 5 to « = 3 which, is second line of
2 k^2'7'2_4 2n K Z e m
Paschen series.
Z
E_=
Sol.9 (A) In lower energylevels kinetic energyis higherand
hence option (D) is correct.
Sol. 2 (D) As radiusof
potential energy and total energy is lesser hence option (A) is orbitofhydrogen atomis givenas h'
r.. =
"
n' X
An^Ke^m
Z
correct.
Sol. 10 (A) In Bohr'sModel the kineticenergyofelectron in orbit is given as
hence option (D) is correct.
1
Sol. 3 (C) As alreadystudiedthat due to differencein masses of nuclei we consider the effect of reduced mass of electron
E=LKZe^
energies ofthe energylevels slightly varyin the twothat cause
2
given as
Sol. 11 (B) InBohr's model thespeed ofthe electron inn"^orbit is given as
1 KZe^ •
hence option (B) is correct.
r„
hence option (A) is correct.
Sol. 4 (B) The total energy of electron in hydrogen atom is
.
2
1 KZe]
and total energy is given as
which is different in hydrogen and deuterium because of which the difference in the spectrum.
2
r
2nKe' v.. =
hence option (B) is correct.
h
fAtomic Physics
41^3.]
Sol. 12 (C) Weuse J_il
47160 ^0
Sol. 22 (C) In case ofelectron we have studied that during collision itcan transfer almost its complete energy to the particle it is colliding hence E^ is the minimum required energy for ionization ofhydrogen atom. In case ofhydrogen and helium
—
ion as helium is more massive than hydrogen it transfers less
v = m
Sol. 13 (A) In
energyduring inelastic collision compared to the case when hydrogen is colliding hence option (C) is correct.
orbit maximum number ofelectrons can be
hencethe possible electrons in energylevels 1, 2, 3 and 4 are 2,8,18 and 32 ofwhich the sum is 2 + 8 +18 + 32 = 60. Hence option (A) is correct.
Sol. 23 (A) The wavelength emitted isinversely proportional to the energy radiated hence option (A) is correct.
Sol. 24 (A) To excite the hydrogen atom the kinetic energy of neutron must be more than 20.4 eV sothatin case ofperfectly
Sol. 14 (D) Because of the energy level differences in the hydrogen atom which explained thattheenergy level difference inelastic collision the energy loss which is half of the total decreases as we go away from nucleus which explained the kinetic energy (10.2 eV) may exite the hydrogen atom soincase hydrogen specturm which other models bythat time couldn't if kientic energyof neutron is less than 20.4 eV then collision must beperfectly elastic hence option (A) is correct. explain hence option (D) is correct. Sol. 15 (D) A quantum number givesthe state of an electron
Sol. 25 (C) As during deexcitation it is emitting six
principle no two electrons can have same set of quantum
emitted energies have less, equal or morethan the excitation energiesthen only option (C) can be correct.
inextranuclear partofanatom soaccording toPauli's exclusion wavelengths that means the atom is excited to «2 =4 and the numbers, hence option (D) is correct.
Sol. 26 (B) Due to the effect of mass of nucleus we consider Sol. 16 (B) From the figure we can see that AE^ =AE^ +AE^ reduce mass of electron in analyzing the energies of energy
hence option (B) is correct as Af = hc/X.
levels of the atoms and due to this the binding energy of deuterium is more thanthatofhydrogen inground state hence
Sol. 17 (D) At room temperature all atoms of hydrogen gas option (B) is correct. Bohr's model is valid for all one electron are inground state inwhich only radiation ofultraviolet region systems (hydrogen like) and some of the lines of Balmer series can be absorbed to excite atoms from « = 1 to higher energy lie in infra red region thats why options (A) and (C) are not levels sovisible light cannot excite any atom in ground state correct. hence option (D) is correct.
Sol. 27 (B) Balmer series is corresponding the deexcitation
Sol. 18 (A) The angular momentum ofelectron in hydrogen
ofelectrons inhydrogen atom to «= 2state hence option (B) is
atom is quantized andaccording toBohr's second postulate for
correct.
energy level it is given as nh/ln so for two successive orbits option (A) is correct.
Sol.28 (C) ByusingRydberg's formula we can calculate the values of wavelengths emitted in above four cases and
Sol. 19 (A) Hydrogen atoms atroom temperature are inground considering same value ofZ for hydrogen and deuterium and state and can only absorb radiation of ultraviolet region to will not consider the effect ofnucleus asin given options we excite from k = 1 to higher energy levels due to which in need tocheck for approximate relations, option (C) iscorrect. absorption specturm of hydrogen gas only Lyman series is obtained hence option (A) is correct.
Sol. 29 (B) As thetotalenergy of electron in hydrogen atom is given as
Sol. 20 (C) aparticle scattering occurs due to Coulomb repulsionof nucleuson incidentaparticle hence option(C) is correct.
^ "
1 KZe^ 2
r„
hence option (B) is correct.
Sol.21 (B) Inany given orbit angular momentum ofelectrons
Sol. 30 (B) 12.leV isthe difference inenergies of« = 1and
are same as given by Bohr's second postulate and energies of electron in anyhydrogen like ion is directly proportional to 2?hence option (B) is correct.
« = 3 forhydrogen atom sothe electron isexcited to « = 3 state hence theangular momentum will become 3/7/271 frran hlln hence theincrease inangular momaitura isgiven inoption (B).